CBSE Class 12 Physics Magnetic Effect of Current Theory and Examples

CBSE Class 12 Physics Magnetic Effect of Current Theory and Examples. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

Moving charges and electric currents produce magnetic fields. This Class 12 Physics notes page on the magnetic effect of current covers Oersted's experiment, magnetic lines of force, the Biot-Savart law, the field due to straight wires, circular coils, arcs and Helmholtz coils, Ampere's circuital law, solenoids and toroids, the force on moving charges and current-carrying conductors, the cyclotron, the moving coil galvanometer and its conversion into an ammeter and a voltmeter, with solved examples.

OERSTED EXPERIMENT (1820)

(a) Oersted found that a magnetic field is established around a current carrying conductor.
(b) Magnetic field exists as long as there is current in the wire.
(c) It is concluded that moving charges produce magnetic field in the surrounding space.
(d) A moving charge produces magnetic as well as electric field, unlike a stationary charge which only produces electric field.
(e) The direction of the magnetic field was found to be changed when direction of current was reversed.

MAGNETIC FIELD OR MAGNETIC INDUCTION OR INTENSITY OF MAGNETIC FIELD (B⃗)

(a) The field produced by flow of current or charge in a conductor is called magnetic field.
(b) This is a vector quantity.
(c) Unit : CGS - Gauss or Maxwell / cm²
MKS - Tesla or Weber / m² or N/Amp-meter
1 Tesla = 1 weber / m² = 10⁴ Gauss = 10⁴ Maxwell / cm²
(d) Magnetic field is shown by magnetic lines of force.

MAGNETIC LINES OF FORCE

(a) These are the imaginary closed curves drawn in magnetic field, which represent the direction of the magnetic field.
(b) The tangent drawn at any point on a line of force shows the direction of magnetic field at that point.

Properties:

(i) The magnetic lines of force always starts from the north pole and following a curved path enter the south pole and reach back the north pole inside the magnet. Thus these are closed curve.

[Figure: Fig (a) - Magnetic lines of force due to current carrying solenoid]

Note : Electric lines of force are not closed curves.
(ii) Two lines of force never intersect each other.
(iii) Larger the number of lines of force at a given point, stronger is the magnetic field at the point.
Note : At poles there are maximum number of lines of force.

BIOT - SAVART LAW

(i) Biot - Savart - a French scientist in 1820 gave a law to study the magnetic field generated by an infinitesimal element of a current carrying conductor at a point P.
(ii) Let the infinitesimal element be of length 'dℓ' and magnetic field at P be dB, then

[Figure: A current carrying conductor XY with current i; a small element dℓ at point C-D of the conductor; r is the line from the element to the point of observation P; θ is the angle between dℓ and r; dB at P is shown perpendicular to the plane of the paper (⊗)]

(a) dB ∝ dℓ
(b) dB ∝ i
(c) dB ∝ sin θ
(d) dB ∝ \( \frac{1}{r^2} \) ⇒ dB ∝ \( \frac{i\,d\ell \sin\theta}{r^2} \) ⇒ dB = \( \frac{\mu_0}{4\pi}\,\frac{i\,d\ell \sin\theta}{r^2} \)
Where \( \frac{\mu_0}{4\pi} \) is a constant
(iii) μ₀ is called the magnetic permeability of vacuum.
Note : ε₀ is called electric permittivity of vacuum i.e. ε₀ refers to electric field while μ₀ refers to magnetic field.
(iv) μ₀ = 4π × 10⁻⁷ Wb/amp-meter or Henry/meter or N/amp²
(v) Vectorially, dB⃗ = \( \frac{\mu_0}{4\pi}\,\frac{i\,(d\vec{\ell} \times \hat{r})}{r^2} \)
⇒ Direction of dB⃗ is perpendicular to both dℓ⃗ and r̂. This is given by right hand screw rule.
(vi) Symbolically. The direction is shown by - ⊗ = inwards (in to the paper), ⊙ = outwards (coming out of the paper normally).
(vii) B⃗ = ∫dB⃗ = \( \frac{\mu_0 i}{4\pi}\int \frac{d\ell \sin\theta}{r^2} \)

Cases :

(a) If sin θ = 0, i.e. θ = 0 or point P lies in the direction of current. Then B = B_min = 0.
(b) If sin θ = 1 i.e. θ = 90º or point P lies perpendicular to the direction of current. Then B = B_max = \( \frac{\mu_0}{4\pi}\,\frac{i\,d\ell}{r^2} \)
(h) Dimension of μ₀ = N / amp² = [MLT⁻² A⁻²]
(i) Dimension of B = N / amp - m = [ML⁰T⁻² A⁻¹]
(j) The direction of magnetic field can be found by any one of the following rules :
(i) Right hand palm Rule
(ii) Right hand thumb Rule.
(iii) Maxwell's Right handed screw Rule
(k) Graphically,

[Figure: Five graphs of B: (i) B versus i is a straight line through the origin; (ii) B versus δℓ is a straight line through the origin; (iii) B versus 1/r² is a straight line through the origin; (iv) B versus θ is a curve that is zero at 0º, has its maximum at 90º and returns to zero at 180º; (v) B versus r is a curve falling steeply and decreasing with r]

(l) Direction of magnetic field : The direction of magnetic field is determined with the help of the following simple laws :
(i) Maxwell's cork screw rule : According to this law if a right handed cork screw is rotated in such a way that it moves forward in the direction of current in the conductor, then the direction of the rotation of the screw will show the direction of lines of force. [fig . (a)]

[Figure: Fig. (a) A cork screw advancing upwards along the current with circular arrows around it showing the direction of the magnetic field B]

(ii) Right hand palm rule : According to this rule if a current carrying conductor is held in the right hand such that the thumb of the hand represents the direction of current flow, then the direction of folding fingers will represent the direction of magnetic lines of force. [Fig. (b)]

[Figure: Fig. (b) A vertical wire carrying current i upward, with concentric circles around it showing the field B and a right hand with the thumb along the current]

(iii) Right hand palm rule of circular currents : According to this rule if the direction of current in circular conducting coil is in the direction of folding fingers of right hand, then the direction of magnetic field will be in the direction of stretched thumb. [Fig. (c)]

[Figure: Fig. (c) A circular current loop with current I, the field lines B passing through the loop along its axis, and a right hand whose fingers curl along the current and thumb points along B]

(m) The magnetic lines of force due to a current carrying element are in the form of concentric circles with their centres at the element.
Note : Biot - Savart law in terms of current density j

dB⃗ = \( \frac{\mu_0}{4\pi}\,\frac{\vec{J} \times \vec{r}}{r^3} \) dv ( ∵ j = \( \frac{i}{A}=\frac{i\,d\ell}{A\,d\ell}=\frac{i\,d\ell}{dv} \) )
where j = current density at any point P of the element dv = volume of element
charge and its velocity
dB⃗ = \( \frac{\mu_0}{4\pi}\,q\,\frac{\vec{v} \times \vec{r}}{r^3} \) where
v = velocity of charge, N = number of charge carriers per unit length of the wire e = charge on charge carrier, q = Neℓ = Total charge in the element dℓ is idℓ = Nveℓ = qv.

REPRESENTATION OF LINES OF MAGNETIC INDUCTION

(a) The lines of magnetic induction due to a straight current carrying conductor are circular.
(b) When the direction of current is upwards i.e. away from plane of paper then lines of magnetic induction are concentric circles in counter clockwise direction.

[Figure: Concentric circles with a dot at the centre and current i out of the paper; the arrows on the circles are counter clockwise]

(c) When the direction of current is down wards i.e. towards the plane of paper then the lines of magnetic induction of concentric circles in clockwise direction.

[Figure: Concentric circles with a cross at the centre and current i into the paper; the arrows on the circles are clockwise and the field is marked B]

Relation between magnetic field and magnetic field intensity H⃗ = B⃗ / μ₀

Ex.1 Dimension of μ₀ is-
(A) MLT⁻² A⁻²   (B) MLT⁻²A⁻²
(C) MLT⁻² A²   (D) MLT² A²

Answer: (A) MLT⁻² A⁻²

1. For a current carrying coil, the field at the centre is B = \( \frac{\mu_0 i}{2R} \) and the force on a length L is F = iLB, so F = \( \frac{\mu_0 i^2 L}{2R} \).
2. Dimensionally, [MLT⁻²] = \( \frac{[\mu_0][A^2][L]}{[L]} \), so [μ₀] = [MLT⁻² A⁻²].
3. Hence, the correct option is (A) MLT⁻² A⁻².

Teacher's Note:
a) Options (A) and (B) are printed identically in the source; the answer is the dimension MLT⁻²A⁻², given as option (A).
b) The relation μ₀ε₀ = 1/c² shows that the dimension of ε₀ is M⁻¹L⁻³A²T⁴.

Note : There is an important relationship between μ₀ and ε₀. This is given by \( \frac{1}{\sqrt{\mu_0 \varepsilon_0}} = c \) or μ₀ε₀ = 1/c²
By this relation, if dimension of one of the μ₀ or ε₀ is known other can be found.
e.g. Dimension of ε₀ = \( \left[ \frac{1}{\mu_0 c^2} \right] \) = \( \frac{1}{MLT^{-2}A^{-2}L^2T^{-2}} \) = M⁻¹ L⁻³ A² T⁴

Ex.2 To obtain maximum intensity of magnetic field at a point, the angle between the position vector of the point relative to the element and the direction of element will be :
(A) 0      (B) π/4
(C) π/2      (D) π

Answer: (C) π/2

1. By the Biot-Savart law, δB = \( \frac{\mu_0}{4\pi}\,\frac{i\,\delta\ell \sin\theta}{r^2} \).
2. δB is maximum when sin θ = 1, that is, θ = 90° = π/2.
3. Then δB = \( \frac{\mu_0 i\,\delta\ell}{4\pi r^2} \) = maximum.
4. Hence, the correct option is (C).

Teacher's Note:
a) The field is zero along the line of the element (θ = 0 or π) and largest in the perpendicular direction.
b) Answer verified from the sin θ dependence of the Biot-Savart law.

MAGNETIC FIELD DUE TO A WIRE OF FINITE LENGTH

B = \( \frac{\mu_0}{4\pi}\,\frac{i(\cos\alpha + \cos\beta)}{r} \) ⊗
= \( \frac{\mu_0}{4\pi}\,\frac{i(\sin\theta_1 + \sin\theta_2)}{r} \) ⊗

[Figure: A vertical wire carrying current I upward from far below (∞) to a point; a perpendicular of length r from the wire to the point P; the angles α and β are measured from the perpendicular ends to the wire ends at P, and θ₁ and θ₂ are the angles the lines from P to the two ends make with the perpendicular. Also a horizontal line O to B with r marked]

Magnetic field due to a semi infinite wire in the formula above, α = π/2 , β = 0 or θ₁ = 0, θ₂ = π/2
⇒ B = \( \frac{\mu_0}{4\pi}\,\frac{i}{r} \) ⊗

Ex.3 A current 1.0 ampere is flowing in the sides of an equilateral triangle of side 4.5 × 10⁻² m. Find the magnetic field at the centroid of the triangle. (Permeability constant μ₀ = 4π × 10⁻⁷ V-s/A-m).

Answer: 4.0 × 10⁻⁵ weber/m²

1. Each side PQ produces at the centroid O a field B = \( \frac{\mu_0}{4\pi}\,\frac{i}{R} \)(sin φ₁ + sin φ₂), where R is the perpendicular distance of PQ from O and φ₁, φ₂ are the angles shown. The field is perpendicular to the plane of the paper, directed downward, and is the same in magnitude and direction for all the three sides, so B = 3 × \( \frac{\mu_0}{4\pi}\,\frac{i}{R} \)(sin φ₁ + sin φ₂).
2. Here i = 1 ampere, φ₁ = φ₂ = 60° and R = (ℓ/2) cot 60° = (1/2)(1/√3)ℓ, where ℓ = 4.5 × 10⁻² m is the side of the triangle.
3. B = \( \frac{3 \times 10^{-7} \times 1.0}{\left(\frac{1}{2} \times 4.5 \times 10^{-2}\right) \times \left(\frac{1}{\sqrt{3}}\right)} \) (\( \frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2} \))
4. = \( \frac{3 \times 10^{-7} \times 2 \times 3}{4.5 \times 10^{-2}} \) = 4.0 × 10⁻⁵ weber/m².
5. Hence, the magnetic field at the centroid is 4.0 × 10⁻⁵ T, perpendicular to the plane of the triangle.

Teacher's Note:
a) The perpendicular distance of the centroid from each side is the inradius, ℓ/(2√3) = 1.3 × 10⁻² m.
b) Check: 3 × 10⁻⁷ × 6/(4.5 × 10⁻²) = 4.0 × 10⁻⁵ T.

Ex.4 A current i flows in a square loop of side 'a'. At the centre of the loop, value of B is-
(A) zero         (B) \( \frac{\mu_0 i}{2\pi a} \)
(C) - \( \frac{2\sqrt{2}\mu_0 i}{2\pi a} \)   (D) \( \frac{2\sqrt{2}\mu_0 i}{\pi a} \)

Answer: (D) \( \frac{2\sqrt{2}\mu_0 i}{\pi a} \)

1. The perpendicular distance d of the centre of the loop from each side is d = a/2.
2. Field due to one side = \( \frac{\mu_0 i}{4\pi d} \)(cos 45° + cos 45°) = \( \frac{\mu_0 i}{4\pi d} \) · √2 = \( \frac{\mu_0 i}{\sqrt{2}\pi a} \) ⊗ (using d = a/2).
3. The fields due to all four sides are in the same direction, so B_total = 4 × \( \frac{\mu_0 i}{\sqrt{2}\pi a} \) = \( \frac{2\sqrt{2}\mu_0 i}{\pi a} \) ⊗.
4. Hence, the correct option is (D).

Teacher's Note:
a) The four fields add because every side makes the current go around the centre in the same sense.
b) If the problem were slightly different, as in a figure where the sides give opposite fields, the answer would have been zero.

MAGNETIC FIELD DUE TO A WIRE OF INFINITE LENGTH

As we know B = \( \frac{\mu_0}{4\pi}\,\frac{i(\sin\theta_1 + \sin\theta_2)}{r} \) [for a wire of finite length]
If θ₁ = 90º and θ₂ = 90º it will be the wire of infinite length
∴ B_infinite = \( \frac{\mu_0}{4\pi}\,\frac{i(1+1)}{r} \) = \( \frac{\mu_0 i}{2\pi r} \)

Ex.5 The magnetic field at a point 50 mm from a long straight line carrying a current of 3A will be-
(A) 0.12 G        (B) 1.2 G
(C) 12 G           (D) 0.012 G

Answer: (A) 0.12 G

1. The field due to a long straight wire is B = \( \frac{\mu_0 i}{2\pi r} \) = \( \frac{4\pi \times 10^{-7} \times 3}{2\pi \times 50 \times 10^{-3}} \).
2. B = 1.20 × 10⁻⁵ Tesla = 0.12 G, since 1 Gauss = 10⁻⁴ Tesla.
3. Hence, the correct option is (A).

Teacher's Note:
a) Convert 50 mm to 0.05 m and use μ₀ = 4π × 10⁻⁷ T m/A.
b) Check: 2 × 10⁻⁷ × 3/0.05 = 1.2 × 10⁻⁵ T.

Ex.6 Two long straight parallel wires carry currents I₁ and I₂ respectively, in the same direction. The distance between the wires is R. The magnetic field at the centre of the two wires will be-
(A) \( \frac{\mu_0 (I_1 - I_2)}{\pi R} \) down wards (If I₁ > I₂)
(B) \( \frac{\mu_0 (I_2 - I_1)}{\pi R} \) upwards (if I₂ > I₁)
(C) \( \frac{\mu_0 (I_1 - I_2)}{\pi R} \) upward
(D) None of the above.

Answer: (A) and (B)

1. The field at a point P between the two wires is B⃗ = B⃗₁ + B⃗₂. The field B₁ (due to current I₁) points downward while B₂ (due to current I₂) points upward.
2. At a distance x from the first wire, B = \( \frac{\mu_0}{2\pi}\left[\frac{I_1}{x} - \frac{I_2}{R-x}\right] \) downwards.
3. At x = R/2, B = \( \frac{\mu_0 (I_1 - I_2)}{\pi R} \) downwards (if I₁ > I₂) or B = \( \frac{\mu_0 (I_2 - I_1)}{\pi R} \) upwards (if I₂ > I₁).
4. Option (C) says upward for I₁ > I₂, which is the wrong direction, and (D) is not true. Hence, the correct options are (A) and (B).

Teacher's Note:
a) For parallel currents the fields between the wires are opposite in direction, so they partly cancel.
b) If I₁ = I₂ and x = R/2, then B = 0. For a point A outside the wires at distances (R + y) and y, the fields add: B = \( \frac{\mu_0}{2\pi}\left[\frac{I_1}{R+y}+\frac{I_2}{y}\right] \) downwards.

MAGNETIC FIELD DUE TO A CIRCULAR CURRENT CARRYING COIL

If ; a = Radius of coil
x = Distance of point P from the centre of coil

[Figure: A circular coil of radius a with centre O' carrying current i; point P on the axis at distance x from the centre; r is the distance from an element dℓ to P; the angle α is between r and the axis; dB from an element at the top and dB from the diametrically opposite element are shown, with the components dB cos α (perpendicular, cancelling) and dB sin α along the axis, which add to 2dB sin α]

B⃗ = \( \left( \frac{\mu_0}{4\pi} \right) \left( \frac{(2\pi a^2) i}{(a^2 + x^2)^{3/2}} \right) \hat{i} \)
If number of turns in coil is 'n', then B ∝ n
⇒ B⃗ = \( \left( \frac{\mu_0}{4\pi} \right) \left( \frac{(2\pi a^2) n i}{(a^2 + x^2)^{3/2}} \right) \hat{i} \)
Where î is the unit vector along x-axis which is the axis of coil in this case.

Special case :

(a) x = 0, i.e. P is centre of coil
B⃗ = \( \frac{\mu_0}{4\pi}\,\frac{2\pi a^2 i}{a^3} \) = \( \frac{\mu_0 i}{2r} \) î
(b) x = ± a,
B = \( \frac{\mu_0}{4\pi}\,\frac{(2\pi a^2) i n}{(2a^2)^{3/2}} \) = \( \frac{\mu_0}{4\pi}\,\frac{2\pi a^2 n}{2\sqrt{2}\,a^3} \) = \( \frac{\mu_0 n i}{4\sqrt{2}\,a} \) ⇒ \( \frac{B_{centre(x=0)}}{B_{(x=\pm a)}} \) = 2√2
(c) x = ± 0.766 R
B = B₀/2, B₀ = B_center

Note : This is the maximum magnetic field, due to coil. i.e. B_max = \( \frac{\mu_0 n i}{2r} \)

POINT OF INFLECTION

(a) These are the points at which rate of change of magnetic field is constant
or, \( \frac{dB}{dx} \) = constant or \( \frac{d^2B}{dx^2} \) = 0
(b) These points are also important because the radius of curvature changes at these point.
(c) In case of circular coil carrying current, x = ± a/2 are called points of inflexion, where 'a' denote the radius of coil.
(d) Graphically, (for circular coil)

[Figure: Graph of B against x for a circular coil: a bell-shaped curve with its maximum B₀ at x = 0, falling on both sides; the points of inflexion A' and A are marked at x = -a/2 and x = a/2]

Ex. 7 A charge of 1 coulomb is placed at one end of a nonconducting rod of length 0.6 m. The rod is rotated in a vertical plane about a horizontal axis passing through the other end of the rod with angular frequency 10⁴ π radian/sec. Find the magnetic field at a point on the axis of rotation at a distance of 0.8 m from the centre of the path.Now half of the charge is removed from one end and placed on the other end. The rod is rotated in a vertical plane about horizontal axis passing through the midpoint of the rod with the same angular frequency. Calculate the magnetic field at a point on the axis at a distance of 0.4 m from the centre of the rod. (μ₀ = 4π × 10⁻⁷ V-s/A-m).

Answer: 1.13 × 10⁻³ tesla and 2.26 × 10⁻³ tesla

1. The rotating rod is a current-loop whose radius a = 0.6 m. The magnetic field due to this current-loop at a point on its axis at a distance x from its centre is B = \( \frac{\mu_0 i a^2}{2(a^2 + x^2)^{3/2}} \) ...(i)
2. Let T be the period of rotation of the rod. Then i = q/T = qω/2π = \( \frac{1\ \text{coulomb} \times 10^4 \pi / \text{sec}}{2\pi} \) = 5 × 10³ amp.
3. Now, a = 0.6 m, x = 0.8 m and μ₀ = 4π × 10⁻⁷ V-s/A-m. Substituting these values in eq. (i): B = \( \frac{(4\pi \times 10^{-7})(5 \times 10^3)(0.6)^2}{2(0.36 + 0.64)^{3/2}} \) = 0.36π × 10⁻³ = 1.13 × 10⁻³ tesla.
4. In the second case the current remains the same because the rotating charge and the angular frequency are the same. However, the radius of the loop becomes half (a = 0.3 m) and the distance x is now 0.4 m.
5. ∴ B = \( \frac{\mu_0 i a^2}{2(a^2 + x^2)^{3/2}} \) = \( \frac{(4\pi \times 10^{-7})(5 \times 10^3)(0.3)^2}{2(0.09 + 0.16)^{3/2}} \) = 0.72π × 10⁻³ = 2.26 × 10⁻³ tesla.
6. Hence, the fields are 1.13 × 10⁻³ T and 2.26 × 10⁻³ T.

Teacher's Note:
a) In the second case the two half-charges at the two ends of the rod move on one circle of radius 0.3 m and together carry the same current as the single charge.
b) Check: 2π × 10⁻³ × 0.18 = 1.13 × 10⁻³ and 2π × 10⁻³ × 0.09/0.25 = 2.26 × 10⁻³.

Ex.8 A circular coil of 0.2 m diameter has 100 turns and carries a current of 0.1 ampere. The intensity of magnetic field at the centre of the coil will be-
(A) 6.28 x 10⁻⁴ N/A.m
(B) 62.8 x 10⁻⁴ N/A.m
(C) 6.28 x 10⁻⁵ N/A.m
(D) 62.8 x 10⁻⁵ N/A.m

Answer: (C) 6.28 x 10⁻⁵ N/A.m

1. The magnetic field at the centre of a current carrying coil having n turns is B = \( \frac{\mu_0 n i}{2r} \) N/A.m, where i is the current in the coil and r is the radius of the coil.
2. Here i = 0.1 A, n = 100 and r = 0.1 m.
3. B = \( \frac{(4\pi \times 10^{-7}) \times 100 \times 0.1}{2 \times 0.1} \) = 6.28 x 10⁻⁵ N/A.m.
4. Hence, the correct option is (C).

Teacher's Note:
a) The printed solution uses n = 1000 and gets 6.28 × 10⁻⁴ (option A), but the question gives 100 turns, so the correct value is 6.28 × 10⁻⁵ (option C).
b) The diameter is 0.2 m, so the radius used in the formula is 0.1 m.

HELMHOLTZ COILS

(a) This is the set-up of two coaxial coils of same radius such that distance between their centers is equal to their radius. Also, same current flows in both in the same direction.
(b) These coils are used to obtain uniform magnetic induction which is obtained between the coils.
(c) Magnitude of the induction between coils is-
B = \( \frac{8\mu_0 n i}{5\sqrt{5}\,R} \) = 0.716 \( \frac{\mu_0 n i}{R} \) = 1.432 B₀, where B₀ = \( \frac{\mu_0 n i}{2R} \)
(d)

[Figure: HELMHOLTZ COIL: two coaxial coils of radius a, with centres O₁ and O₂ a distance a apart, the point O midway, and the supply terminals + and -. Below it a graph of B against x shows the two bell-shaped curves of the coils (centred at x = -a/2 and x = a/2) whose sum, drawn dotted, is flat in the middle: the uniform field region around x = 0]

MAGNETIC FIELD DUE TO AN ARC MAKING AN ANGLE α AT THE CENTRE

For fig (a) B = \( \frac{\mu_0 i \alpha}{4\pi a} \), normal to the plane of paper downwards.

[Figure: Fig (a): an arc PQ of radius a subtending the angle α at the centre O, with current i along the arc from P to Q (anticlockwise sense); O is shown with a cross mark (field into the paper)]

Fig (b) B = \( \frac{\mu_0 i \alpha}{4\pi a} \), normal to the plane of paper upwards.

[Figure: Fig (b): the same arc with the current reversed (from Q to P), and O is shown with a dot (field out of the paper)]

[Figure: Two concentric circular coils with centre O and radii R₁ (inner) and R₂ (outer) carrying currents I₁ and I₂. In figure (a) the currents I₁ and I₂ flow in the same sense; in figure (b) they flow in opposite senses]

Ex.9 Consider two concentric coils, radii R₁ and R₂ number of turns N₁, N₂ carrying currents I₁ and I₂ as shown in fig. (a) and (b). The magnetic field at the centre if I₁ = I₂ = I and N₁ = N₂ = N in these figures will be respectively-
(A) \( \frac{\mu_0 N I (R_1 + R_2)}{2 R_1 R_2} \), \( \frac{\mu_0 N I (R_2 - R_1)}{2 R_1 R_2} \)
(B) 0, \( \frac{\mu_0 N I (R_2 - R_1)}{2 R_1 R_2} \)
(C) \( \frac{\mu_0 N I (R_2 - R_1)}{2 R_1 R_2} \), 0
(D) 0, 0

Answer: (A)

1. (a) In fig. (a) the fields due to both coils are in the same direction, so B = \( \frac{\mu_0 N_1 I_1}{2R_1} + \frac{\mu_0 N_2 I_2}{2R_2} \). For I₁ = I₂ = I and N₁ = N₂ = N, B = \( \frac{\mu_0 N I (R_1 + R_2)}{2 R_1 R_2} \).
2. (b) In fig. (b) the fields due to the two coils are in opposite direction, so the total field is B = \( \frac{\mu_0 N_1 I_1}{2R_1} - \frac{\mu_0 N_2 I_2}{2R_2} \). For the same values, B = \( \frac{\mu_0 N I (R_2 - R_1)}{2 R_1 R_2} \).
3. Hence, the correct option is (A).

Teacher's Note:
a) The fields add when the two currents circulate in the same sense and subtract when they circulate in opposite senses.
b) The sign in part (b) assumes R₂ greater than R₁, so that the smaller coil dominates.

[Figure: Two identical circular coils, Coil 1 and Coil 2, placed in perpendicular planes with a common centre O, with the X and Y axes and the resultant field B making an angle θ with the X axis]

Ex. 10 For the arrangement of fig the magnetic field at the center O will be-
(A) √2 \( \left( \frac{\mu_0 N I}{2a} \right) \)
(B) \( \frac{\mu_0 N I}{2\sqrt{2}\,a} \)
(C) \( \frac{\mu_0 N I}{2} \)
(D) \( \frac{\mu_0 N I}{2a} \)

Answer: (A) √2 \( \left( \frac{\mu_0 N I}{2a} \right) \)

1. The two coils are perpendicular to each other. Coil 1 produces its field along the X axis and coil 2 produces its field along the Y axis.
2. The resultant field is B = √(B₁² + B₂²), making an angle θ with tan θ = B₂/B₁.
3. For identical coils, B = √2 \( \left( \frac{\mu_0 N I}{2a} \right) \) and θ = 45°.
4. Hence, the correct option is (A).

Teacher's Note:
a) Perpendicular fields are added by the Pythagoras rule, not by simple addition.
b) Each coil gives the field μ₀NI/2a at the centre, since each has the same radius a and the same turns.

MAGNETIC FIELD DUE TO VARIOUS GEOMETRIES

Geometry and Magnetic field

(a) [Figure: Two concentric circular coils of radii R₁ and R₂ carrying currents I₁ and I₂]
B = \( \left[ \frac{\mu_0 I_1}{2R_1} + \frac{\mu_0 I_2}{2R_2} \right] \) ⊙

(b) [Figure: A straight wire carrying current i, with the point P on the line of the wire, ahead of its end]
B = 0

(c) [Figure: A semicircular arc of radius a with centre P, with straight leads on both sides along the diameter line; the current i flows along the leads and the arc]
B = \( \frac{\mu_0}{4\pi}\,\frac{i \cdot \pi}{a} \) = \( \frac{\mu_0 i}{4a} \) ⊗

(d) [Figure: Two concentric semicircular arcs of radii a₁ and a₂ with centre P, joined by straight segments on the same line (points A, B, C, D marked); the current i flows through the arrangement]
B = \( \frac{\mu_0}{4\pi} \)(πi)\( \left( \frac{1}{a_1} - \frac{1}{a_2} \right) \) ⊗

(e) [Figure: Two concentric semicircular arcs of radii a₁ and a₂ with centre P, joined by straight segments on the same line (points A, D, C, B marked); the current i flows through the arrangement]
B = \( \frac{\mu_0 \pi i}{4\pi}\left( \frac{1}{a_1} + \frac{1}{a_2} \right) \) ⊙

(f) [Figure: A circular loop of radius a with centre P, touching a long straight wire B-C-D at the point C; the current i flows in the wire and in the loop]
B = \( \frac{\mu_0 i}{2a}\left( 1 - \frac{1}{\pi} \right) \)

(g) [Figure: A circular loop of radius a with centre P, touching a long straight wire A-B; the current i flows in the wire and in the loop]
B = \( \frac{\mu_0 i}{2a}\left( 1 + \frac{1}{\pi} \right) \) ⊙

(h) [Figure: Two circular arcs of radii a₁ and a₂ with the common centre P, joined by straight segments; θ is the angle between the straight segments; the current is i]
B = \( \frac{\mu_0 i}{4\pi}\left\{ \frac{2\pi - \theta}{a_1} + \frac{\theta}{a_2} \right\} \) ⊗

(i) [Figure: Two perpendicular circular coils, coil 1 of radius a₁ and coil 2 of radius a₂, with the common centre P, carrying currents i₁ and i₂]
B₁ = \( \frac{\mu_0}{4\pi}\,\frac{2\pi i_1}{a_1} \) k̂, B₂ = \( \frac{\mu_0}{4\pi}\,\frac{2\pi i_2}{a_2} \) ĵ, B = √(B₁² + B₂²) = \( \frac{\mu_0}{2}\sqrt{\frac{i_1^2}{a_1^2} + \frac{i_2^2}{a_2^2}} \)

(j) [Figure: A circular loop with centre P; the current i enters at A and leaves at C, dividing into i₁ and i₂ in the two arcs]
B = 0

Note : Check out these formulae yourself

Special Note :

1. If in a symmetrical geometry, current enters from one end and exit from the other, then magnetic field at the center is zero.

[Figure: Five symmetrical conductors with the current entering at one point and leaving at the opposite point, each with B at the centre C equal to zero: (1) a circle, (2) a rectangle with the current entering at one corner and leaving at the opposite corner, (3) a square, (4) a cube with the current entering at one corner and leaving at the diagonally opposite corner, (5) a triangle frame]

Note : in all the above cases, B_C = 0

AMPERE'S LAW

(a) Line integral of the magnetic field B⃗ around any closed curve is equal to μ₀ times the net current i threading through the area enclosed by the curve i.e.
∮ B⃗ · dℓ⃗ = μ₀ Σi = μ₀ (i₁ + i₂ + i₃)

[Figure: A closed curve with three currents i₁, i₂ and i₃ passing through it, a small element dℓ on the curve and the field B making the angle θ with it]

(b) The line integral of magnetic field around a closed path is equal to μ₀ times the total current enclosed by the closed path.
∮ B⃗ · dℓ⃗ = μ₀ Σi = μ₀ Σ (-i₁ + i₂ - i₃)

[Figure: A closed path with currents i₁, i₂ and i₃ passing through it in different directions (i₁ and i₃ in the opposite sense to i₂), with the element dℓ and the angle θ]

(c) The Ampere's Law in magnetostatic is equivalent to Gauss's law in electrostatics.
(d) When the direction of current is away from the observer then the direction of closed path is clockwise.

[Figure: A current i directed away from the observer through a loop, with the closed path marked clockwise]

(e) When the direction of current is towards the observer then the direction of closed path is anticlockwise.

[Figure: A current i directed towards the observer through a loop, with the closed path marked anticlockwise]

Difference between Biot - Savart law and Ampere's law

Biot - Savart - Law
(a) This law is valid for asymmetrical current distributions
(b) This law is the differential form of B⃗ or H⃗
(c) This law is based only on the principle of magnetism
Ampere's Law
(a) This law is valid for symmetrical current distributions.
(b) Basically this law is the integral form of B⃗ or H⃗
(c) This law is based on the principle of electromagnetism.

Application of Ampere's law

(a) Magnetic field of a long straight wire :
Let op = r

[Figure: A long straight wire carrying current i upward through the centre O of a circular path of radius r in the plane of the paper, with a point P on the circle; the symbols ⊙ and ⊗ mark the wire crossing the plane]

By Ampere's Law, ∮ B⃗ · dℓ⃗ = μ₀ i
Since B is constant through out the circle, Let it be B.
B ∮ dℓ = μ₀ i
B · 2πr = μ₀ i ⇒ B = \( \frac{\mu_0 i}{2\pi r} \)
⇒ B ∝ \( \frac{1}{r} \) hence B_∞ = 0

[Figure: Cross-section of a hollow conducting pipe (shaded ring) with the point P at the centre of the hollow region, the point Q in the conducting material at the lower left, and the point R outside the pipe above it]

Ex.11 Current is flowing through a conducting hollow pipe whose area of cross-section is shown as. The value of magnetic induction will be zero at-
(A) Point P, Q and R
(B) Point R but not at P and Q
(C) Q but not at P and R
(D) P but not at Q and R

Answer: (D) P but not at Q and R

1. Apply Ampere's law at P, Q and R respectively.
2. The circle through P, drawn inside the hollow part of the pipe, encloses no current, so ∮B · dℓ = 0 and the magnetic induction at P is zero.
3. The circles through Q (in the material of the pipe) and R (outside the pipe) enclose current, so the magnetic induction at Q and R is non-zero.
4. Hence, the correct option is (D).

Teacher's Note:
a) A field is zero inside a hollow conductor only if the current has cylindrical symmetry.
b) The field inside the hollow region of a long current-carrying pipe is zero, just as in a cylindrical shell.

Magnetic field due to a current carrying cylinder

(A) Cylindrical shell :
B₁ = 0 (inside the shell, r < R)
B₂ = \( \frac{\mu_0 I}{2\pi R} \) (on the surface, r = R)
B₃ = \( \frac{\mu_0 I}{2\pi r} \) (outside the shell, r > R)

[Figure: A graph of B against r for a cylindrical shell of radius R: B is zero up to r = R, then it falls as the curve of 1/r for r greater than R. Next to it a hollow cylinder of radius R carrying current I upward with three points marked 1 (inside), 2 (on the surface) and 3 (outside)]

(B) Rigid Cylinder :
B₁ = \( \frac{\mu_0 I r}{2\pi R^2} \) (inside, r < R)
B₂ = \( \frac{\mu_0 I}{2\pi R} \) , B₃ = \( \frac{\mu_0 I}{2\pi r} \) (at r = R and r > R)

[Figure: A graph of B against r for a solid cylinder: for r less than R the line rises through the origin (B ∝ r), the peak is at r = R, and for r greater than R it falls as the curve B ∝ 1/r. Next to it a solid cylinder]

(c) Magnetic field due to a charged ring
Let the frequency of ring = n and charge on ring = q. Then (a) current I = nq

[Figure: A charged ring with the point at its centre marked and the axis through the centre]

(b) Magnetic field at the centre
= \( \frac{\mu_0 I}{2R} \) = \( \frac{\mu_0 n q}{2R} \) = \( \frac{\mu_0 \omega q}{2\pi R} \) (ω = 2πn)
(c) Magnetic moment = M = i A = (qn) (πR²)
⇒ M = πqnR² or M = \( \frac{q\omega R^2}{2} \)

SOLENOID

(a) It is a long cylindrical helix made by winding closely a large number of turns of insulated copper wire over a card board or china clay.
(b) The winding or wire is uniform.
(c) A magnetic field is produced around and within the solenoid. The magnetic field within the solenoid is uniform and parallel to the axis of Solenoid.
(d) Magnetic field outside the solenoid is negligible.
(e) For an ideal solenoid, length is very-very greater than it's radius.
(f) If i is the current flowing in solenoid then magnetic field inside the solenoid
B = μ₀ n i
Where n = number of turns in unit length

[Figure: A solenoid with dashed field lines running parallel to the axis inside it and spreading out at both ends; B = 0 is marked outside]

(g) If length of solenoid = L(>> r) and number of turns = N, then
n = \( \frac{N}{L} \) ⇒ B = μ₀ \( \frac{N}{L} \) i
(h) Magnetic field at the center is twice the magnetic field at an end of the solenoid i.e.
B_end = \( \frac{\mu_0 n i}{2} \)
(i) The magnetic field at any point P due to a solenoid of finite length is given by

[Figure: A solenoid with a point P on its axis; the lines from P to the two ends of the solenoid make the angles α₁ and α₂ with the axis]

B = \( \frac{1}{2} \) μ₀ ni (cos α₂ - cos α₁)

Ex. 12 An air-solenoid has 500 turns of wire in its 40 cm length. If the current in the wire be 1.0 ampere then the magnetic field on the axis inside the solenoid will be-
(A) 15.7 gauss      (B) 1.57 gauss
(C) 0.157 gauss    (D) 0.0157 gauss

Answer: (A) 15.7 gauss

1. The magnetic field inside (near centre) a current carrying solenoid having n turns per unit length is given by B = μ₀ ni newton/(ampere-meter), where i (ampere) is the current in the solenoid and μ₀ = 4π × 10⁻⁷ newton/ampere².
2. Here n = 500/0.40 = 1250 per meter and i = 1.0 amp.
3. ∴ B = (4 × 3.14 × 10⁻⁷) × 1250 × 1.0 = 15.7 × 10⁻⁴ newton/(ampere-meter) = 15.7 gauss.
4. Hence, the correct option is (A).

Teacher's Note:
a) 1 gauss = 10⁻⁴ tesla, so 1.57 × 10⁻³ T = 15.7 gauss.
b) The answer applies near the centre of a long solenoid, where the field is uniform.

Ex.13 A solenoid of length 0.2m has 500 turns on it. If 8.71 × 10⁻⁶ W/m² be the magnetic field at an end of the solenoid, then the current flowing in the solenoid will be -
(1) \( \frac{0.174}{\pi} \) amp.    (2) \( \frac{174}{\pi} \) amp.
(3) \( \frac{17.4}{\pi} \) amp.    (4) \( \frac{174}{\pi} \) amp.

Answer: (1) \( \frac{0.174}{\pi} \) amp.

1. We know, B_end = \( \frac{\mu_0 n i}{2} \). Here n = 500/0.2 = 2500/metre, i = ? and B_end = 8.71 × 10⁻⁶.
2. ∴ i = \( \frac{2B_{end}}{\mu_0 n} \) = \( \frac{2 \times 8.71 \times 10^{-6}}{4\pi \times 10^{-7} \times 2500} \) = \( \frac{0.01742}{\pi} \) amp for the value printed in the question.
3. The listed options, with the printed key, correspond to B_end = 8.71 × 10⁻⁵ W/m², which gives i = \( \frac{2 \times 8.71 \times 10^{-5}}{\pi \times 10^{-3}} \) = \( \frac{0.1742}{\pi} \) amp.
4. Hence, the answer is option (1), 0.174/π amp.

Teacher's Note:
a) With B_end = 8.71 × 10⁻⁶ exactly as printed, i = 0.0174/π A = 5.5 × 10⁻³ A, which is not among the options; the options and the printed answer correspond to B_end = 8.71 × 10⁻⁵ W/m².
b) Options (2) and (4) are printed identically in the source.

Ex.14 A solenoid of length 0.4 m and diameter 0.6 m consists of a single layer of 1000 turns of fine wire carrying a current of 5.0 × 10⁻³ ampere. The magnetic field on the axis at the middle and at the ends of the solenoid will be wb/m² - (Given μ₀ = 4π × 10⁻⁷ V-s/A-m).
(A) 8.71 × 10⁻⁶, 3.14 x 10⁻⁷
(B) 8.71 × 10⁻⁶, 6.28 x 10⁻⁶
(C) 8.71 × 10⁻⁶, 0
(D) 0, 8.71 × 10⁻⁶

Answer: (B) 8.71 × 10⁻⁶, 6.28 x 10⁻⁶

1. The magnetic field at the middle point P on the axis of a current carrying solenoid is given by B = \( \frac{1}{2} \) μ₀ ni (cos θ₁ - cos θ₂), where n is number of turns per meter and i is current (in ampere). Angles θ₁ and θ₂ are as shown in the figure.
2. Here n = \( \frac{1000}{0.4} \) = 2500 per meter and i = 5.0 × 10⁻³ ampere. At the middle the radius is 0.3 m and the half-length is 0.2 m, so cos θ₁ = \( \frac{0.2}{\sqrt{(0.3)^2 + (0.2)^2}} \) = 0.5547 and cos θ₂ = - 0.5547.
3. ∴ B = \( \frac{1}{2} \) × (4 × 3.14 × 10⁻⁷) × 2500 × (5.0 × 10⁻³) × (0.5547 + 0.5547) = 8.71 × 10⁻⁶ weber/m².
4. At the end of the solenoid, cos θ₁' = \( \frac{0.4}{\sqrt{(0.3)^2 + (0.4)^2}} \) = \( \frac{0.4}{0.5} \) = 0.8 and cos θ₂' = cos 90º = 0.
5. ∴ B = \( \frac{1}{2} \) × (4 × 3.14 × 10⁻⁷) × 2500 × (5.0 × 10⁻³) × 0.8 = 6.28 × 10⁻⁶ weber/m².
6. Hence, the correct option is (B).

Teacher's Note:
a) The diameter 0.6 m is larger than the length 0.4 m, so the solenoid is not long and the end value is not exactly half of the middle value.
b) Check: 0.5 × 4π × 10⁻⁷ × 2500 × 5 × 10⁻³ = 7.85 × 10⁻⁶, multiplied by 1.109 and 0.8 gives 8.71 × 10⁻⁶ and 6.28 × 10⁻⁶.

Ex.15 A solenoid of length 0.4 m and having 500 turns of wire carries a current of 3 amp. a thin coil having 10 turns of wire and of radius 0.01 m carries a current of 0.4 amp. The torque required to hold the coil in the middle of the solenoid with its axis perpendicular to the axis of the solenoid.
(μ₀ = 4π × 10⁻⁷ V-s/A-m) will be (in Nm) -
(A) 59.2 × 10⁻⁶      (B) 5.92 × 10⁻⁶
(C) 0.592 × -10⁻⁶   (D) 0.592 × 10⁻⁴.

Answer: (B) 5.92 × 10⁻⁶

1. The magnetic field at the centre of the solenoid is B = μ₀ ni = (4π × 10⁻⁷) × (500/0.4) × 3 N/A.m.
2. The torque acting on a current-carrying coil having N turns placed perpendicular to the axis at the centre of the solenoid is τ = BiNA = (4π × 10⁻⁷) × (500/0.4) × 3 × 0.4 × 10 × π (0.01)².
3. = 6π² × 10⁻⁷ = 5.92 × 10⁻⁶ N.m.
4. Hence, the correct option is (B).

Teacher's Note:
a) The coil's magnetic moment is perpendicular to the field, so the torque takes its maximum value MB.
b) Option (C) is printed with a stray minus sign in the source.

TOROID

(a) This is a solenoid bent round in the form of a closed ring and carrying a current.
(b) Magnetic field within the toroid is uniform and outside it is zero.
(c) Magnetic field inside the toroid
B = μ₀ ni = μ₀ \( \frac{N}{2\pi R} \) i
Where N = Total number of turns.
R = Radius of toroid.

[Figure: A toroid of radius R wound with turns of wire, with the circular field lines (B lines) inside it; B = 0 is marked inside the hole and outside the toroid]

(d) If magnetic permeability of material of toroid, μ then
B = μni
Here, μ = μ₀ μ_r
⇒ B = μ₀ μ_r ni
Where μ_r = Relative permeability of medium.

MAGNETIC MOMENT OF A CURRENT CARRYING COIL

(a) All current carrying conductors behave like a magnet.
(b) If current in a coil is flowing anticlockwise, then that plane of the coil behaves as north pole. Similarly, if current is flowing clock wise, plane behaves as south pole as shown in the figure.

[Figure: Two circular coils: one with an anticlockwise current, whose face acts as the north pole, and one with a clockwise current, whose face acts as the south pole]

(c) Magnetic moment (M) is given by the product of current and effective area of the coil i.e. M = I A.
(d) Magnetic moment of a coil of radius r and having 'n' turns is given by M = n(πr²) I
(e) Unit of magnetic moment : Ampere - meter²
Dimension : [M⁰ L² T⁰ A¹]
(f) Magnetic moment is a vector quantity. It's direction is given as south pole to north pole of the magnet.
(g) Magnetic field at the center of coil can also be written as
B = \( \frac{\mu_0 I}{2R} \) = \( \frac{\mu_0 M}{2\pi R^3} \)
Where M = Magnetic moment.

Ex.16 The average radius of a toroid made from a paramagnetic material is 0.1 m and it has 500 turns. If it carries 0.5 ampere current, then the magnetic field inside it will be :
(A) 5 × 10⁻⁴ tesla     (B) 5 × 10⁻³ tesla
(C) 5 × 10⁻² tesla     (D) 2 × 10⁻³ tesla

Answer: (A) 5 × 10⁻⁴ tesla

1. The field inside a toroid is B₀ = μ₀ \( \frac{Ni}{2\pi R} \).
2. B₀ = \( \frac{4\pi \times 10^{-7} \times 500 \times 0.5}{2\pi \times 0.1} \) = 5 × 10⁻⁴ tesla.
3. Hence, the correct option is (A).

Teacher's Note:
a) For a paramagnetic material the relative permeability is very close to 1, so the vacuum value is used.
b) Check: 2 × 10⁻⁷ × 250/0.1 = 5 × 10⁻⁴ T.

Ex.17 If an electron is moving with velocity v in an orbit of radius r in a hydrogen atom, then the equivalent magnetic moment will be :
(A) \( \frac{\mu_0 e}{2r} \)        (B) \( \frac{ev}{r^2} \)
(C) \( \frac{ev \times 10^{-7}}{r^3} \)   (D) \( \frac{evr}{2} \)

Answer: (D) \( \frac{evr}{2} \)

1. The equivalent magnetic moment is M = iA = ef (πr²), where the frequency f = \( \frac{v}{2\pi r} \).
2. ∴ M = \( \frac{ev}{2\pi r} \) πr² = \( \frac{evr}{2} \).
3. Hence, the correct option is (D).

Teacher's Note:
a) A revolving electron is equivalent to a current loop with i = ef.
b) The magnetic moment of an orbiting charge is half of its charge times velocity times radius.

Ex.18 On account of the orbital motion of an electron its magnetic moment will be (h - Plank constant, e-electronic charge and m-mass of an electron)
(A) \( \frac{eh}{4\pi m} \)       (B) \( \frac{h}{4\pi m} \)
(C) \( \frac{eh}{2\pi} \)       (D) \( \frac{eh}{2\pi m} \)

Answer: (A) \( \frac{eh}{4\pi m} \)

1. The magnetic moment on account of the orbital motion of an electron is M = \( \frac{evr}{2} \).
2. From Bohr's quantum condition, mvr = \( \frac{nh}{2\pi} \), and for n = 1, vr = \( \frac{h}{2\pi m} \).
3. ∴ M = \( \frac{e}{2}\cdot\frac{h}{2\pi m} \) = \( \frac{eh}{4\pi m} \).
4. Hence, the correct option is (A).

Teacher's Note:
a) This is the Bohr magneton, 9.27 × 10⁻²⁴ A m².
b) Use n = 1 for the first (ground-state) orbit.

MOTION OF A CHARGED PARTICLE IN A MAGNETIC FIELDS

(a) If a particle carrying a positive charge q and moving with velocity V⃗ enters a magnetic field B⃗ then it experiences a force F⃗ given by
F⃗ = q(V⃗ × B⃗) , |F| = qVB sin θ
where θ is the angle between V⃗ and B⃗ and direction of F is perpendicular to the plane formed by V⃗ and B⃗.
(b) Direction of F is also given by right hand palm rule as shown i.e. B⃗ is along the thumb Then F⃗ has the direction upward perpendicular to the plane of palm. In case the charge is negative, direction is downward perpendicular to plane of palm.

[Figure: Two sketches of a charge q with velocity v in a field B along the x-axis in the plane: in the first the angle between v and B is 90º and the force F points along z perpendicular to the plane; in the second the angle is θ and the component v sin θ is marked, with F along z]

(c) If a charge does not experience any force even if kept in a magnetic field, then the possibilities are-
(i) q = 0 i.e. particle is uncharged
(ii) V = 0 i.e. particle is stationery
(iii) θ = 0 i.e. particle is moving in the direction of magnetic field.
(d) The maximum force experienced by a particle in a magnetic field B is given by
F_max = qVB
i.e. θ = \( \frac{\pi}{2} \) or particle is moving in a direction perpendicular to the magnetic field.
(e) Since the force on the particle is perpendicular to the direction of motion or displacement, No work is done by the magnetic field on a moving charge.
There is no change in kinetic energy (KE) of the particle in a magnetic field.
There is no change in speed or magnitude of velocity, although direction keeps on changing continuously.
There is no change in momentum of the particle although its direction also keeps on changing continuously.
(f) (Imp.) If a particle enters a magnetic field normally or perpendicular to the magnetic field, then it starts moving in a circular orbit. Some facts of this circular motion are given below.
(i) The point at which it enters the magnetic field lies on the circumference. (Most of us confuse it with the center of the orbit).
(ii) Centripetal force is provided by the magnetic force i.e.
\( \frac{mv^2}{R} \) = qvB ⇒ R = \( \frac{mv}{qB} \) = \( \frac{p}{qB} \) = \( \frac{\sqrt{2mE}}{qB} \)
where P = mv = linear momentum.
E = P²/2m = kinetic energy
(iii) Kinetic Energy
E = \( \frac{1}{2} \)mv² = \( \frac{R^2 q^2 B^2}{2m} \)
(iv) Time - period (T) of the particle
T = \( \frac{2\pi R}{v} \) = \( \frac{2\pi m}{qB} \)

Note :
That time period does not depend upon the velocity with which the particle enters the magnetic field.
Note (Imp) :
When a charged particle enters normally to electric field then its path becomes parabolic while in magnetic field circular.
(g) If the particle enters at an angle θ then the particle will move in to a helical path, Why ?
Velocity 'V' of the particle may be resolved in to two components. One along the magnetic field and the other perpendicular to the magnetic field.
V₁ = V cos θ and V₂ = V sin θ.
V₂ gives a circular path to the particle while V₁ gives a linear path. The resultant is a helical path.
(i) The radius of circular path = \( \frac{mV\sin\theta}{qB} \)
(ii) Time - period does not depend on velocity and hence T = \( \frac{2\pi m}{qB} \)
(iii) P = pitch = distance between two circles = (V cos θ) \( \left( \frac{2\pi m}{qB} \right) \)

Ex.19 An α particle travels at an angle of 30º to a magnetic field 0.8 T with a velocity of 10⁵ m/s. The magnitude of force will be-
(A) 12.8 × 10⁻¹⁴ N    (B) (1.28)√3 ×10⁻⁴ N
(C) 1.28 × 10⁻⁴ N     (D) (12.8)√3 ×10⁻⁴ N

Answer: (C)

1. F = qvB sin θ = 2 × 1.6 × 10⁻¹⁹ × 10⁵ × 0.8 × (1/2).
2. = 1.28 × 10⁻¹⁴ N, since the charge on an α particle = 2e.
3. The value 1.28 × 10⁻¹⁴ N is the one that the key marks as option (C).

Teacher's Note:
a) The calculated force is 1.28 × 10⁻¹⁴ N; option (C) is printed as 1.28 × 10⁻⁴ N, which is a misprint of the exponent.
b) A force of this size is very small because the charge is only 3.2 × 10⁻¹⁹ C.

[Figure: A beam of protons moving with the velocity v through a region of the field directed downward (shown by crosses); the force F is towards the right]

Ex.20 A beam of protons is moving horizontally towards you. As it approaches, it passes through a magnetic field directed downward. The beam deflects-
(A) to your left side
(B) to your right side
(C) does not deflect
(D) nothing can be said

Answer: (B) to your right side

1. The direction of F⃗ is along (V⃗ × B⃗) for a positive charge, and here V⃗ is towards you and B⃗ is downward.
2. V⃗ × B⃗ is towards the right of the observer.
3. Thus the beam deflects to your right side.
4. Hence, the correct option is (B).

Teacher's Note:
a) If instead of a proton beam, an electron beam moving towards you is considered, it will be deflected towards left because the charge is negative.
b) Think: the force is at right angles to both the velocity and the field.

[Figure: Four particles (1), (2), (3) and (4) enter a magnetic field acting upwards; the figure is not available in the source]

Ex.21 Particles (1), (2) and (4) enters into the magnetic field acting upwards as shown in the fig. Identify the particles.

Answer: (1) positive, (2) neutral, (3) negative, (4) positive

1. (1) is positive charge as the force is acting in upward direction i.e. F = evB.
2. (2) is neutral particle, there is no force on it, i.e. F = 0.
3. (3) is negative charge, as the force is acting downwards i.e. F = - evB.
4. The path of particle (4) indicates that it must be positively charged particle.

Teacher's Note:
a) The question text prints (1), (2) and (4), while the solution discusses particles (1) to (4); all four are identified here.
b) A neutral particle goes straight; the other two bend in opposite senses according to the sign of the charge.

Ex. 22 If a particle moves in a circular path in clock wise direction after entering in to a downward vertical magnetic field. The charge on the particle is-
(A) positive
(B) negative
(C) nothing can be said
(D) Neutral

Answer: (B) negative

1. The particle is moving clockwise, which shows that the force on the particle is opposite to that given by the right hand palm rule or Fleming left hand rule.
2. These two laws are used for positive charge. Here since the laws are disobeyed, we can say that the charge is negative.
3. Hence, the correct option is (B).

Teacher's Note:
a) If the magnetic field were vertically upward, the sense of rotation for the same charge would reverse.
b) The question is printed as "The change on the particle is"; the intended meaning is the charge on the particle.

Ex.23 In the example above, after how much time, particle comes to the starting point for the first time. (mass of particle = m)
(A) Never          (B) \( \frac{2\pi m}{qB} \)
(C) \( \frac{2\pi m}{3qB} \)   (D) It will leave the circular path before coming to the point

Answer: (B) \( \frac{2\pi m}{qB} \)

1. The starting point lies on the circumference of the circular path, hence the particle will come back after one time period T.
2. T = \( \frac{2\pi m}{qB} \).
3. Hence, the correct option is (B).

Teacher's Note:
a) The period does not depend on the speed or the radius of the orbit.
b) The field is assumed to extend over the full circle, so the particle completes it.

FORCE ON A CURRENT CARRYING CONDUCTOR IN A MAGNETIC FIELD

(a) Consider an electron inside the conductor with velocity V⃗_d (V⃗_d is antiparallel to the direction of current). Force on this electron
= e (V⃗_d × B⃗) magnitude of force = eV_dB sin (π - θ) = eV_dB sin θ.
If length of the conductor is 'ℓ' number of electron per unit volume = 'n'
Area of cross section = 'A'
Total number of electrons = n(Aℓ)
Total force = n(Aℓ) eV_dB sin θ
= (ne V_dA) ℓB sin θ
= iℓB sin θ (i = ne V_dA)
(b) Vectorially, F⃗ = i (ℓ⃗ × B⃗)
Where direction of ℓ⃗ is taken along the direction of flow of current. Direction of force F⃗ is perpendicular to the plane formed by ℓ⃗ and B⃗. This is given by Fleming's left hand rule. i.e. If the forefinger, the middle finger and thumb of the left hand are stretched mutually at right angles to one another such that direction of magnetic field B⃗ → along the fore finger, direction of current i → along middle finger. The force F⃗ will be in the direction of thumb.

[Figure: A straight conductor carrying the current i placed at an angle θ to a set of parallel field lines B, with the force F perpendicular to the plane of the conductor and the field]

(c) |F| = i ℓ B sin θ
i.e. if θ = 0, 180º or wire is kept parallel or antiparallel to the direction of magnetic then force on the conductor is zero.

[Figure: A charged particle path in a magnetic field B drawn as a series of loops advancing to the right (helical motion)]

(d) F will be maximum when θ = 90º
F_max = i ℓ B.

Ex.24 A current of 2.0 amp. is flowing through a wire of length 50 cm. If this wire be placed at an angle of 60º with the direction of a uniform magnetic field of 5.0 × 10⁻⁴ N/A.m. the force on the wire will be-
(A) 4.33 × 10⁻⁴ N     (B) 2.50 × 10⁻⁴ N
(C) 5.0 × 10⁻⁴ N     (D) 2.33 × 10⁻⁴ N

Answer: (A) 4.33 × 10⁻⁴ N

1. The magnetic force on a current carrying wire of length L, placed in a magnetic field B at an angle θ with the field is given by F = i ℓ B sin θ.
2. Here B = 5.0 × 10⁻⁴ N/A.M, i = 2.0 A, ℓ = 50 cm = 0.50 m and θ = 60º.
3. F = 2.0 × 0.50 × (5.0 × 10⁻⁴) × sin 60º = 4.33 × 10⁻⁴ N.
4. According to the Fleming's left - hand rule, this force will act perpendicular to both the wire and the magnetic field.
5. Hence, the correct option is (A).

Teacher's Note:
a) sin 60° = 0.866, so the force is 0.866 times the maximum value of 5.0 × 10⁻⁴ N.
b) The units N/A.m for B are the same as tesla.

[Figure: A rectangular coil PQRS placed in the magnetic field with its side PQ parallel to the field; the figure is not available in the source]

Ex.25 A rectangular coil is placed in magnetic field in such a way that its side PQ is parallel to the field. The moments of the couples acting on the four sides of coil for current of 2.0 A in the coil and B = 200 G will be-
(A) 1 × 10⁻² N.m     (B) 4.0 × 10⁻³ N.m.
(C) 3 × 10⁻³ N.m.     (D) 2 × 10⁻² N.m.

Answer: (B) 4.0 × 10⁻³ N.m.

1. We know, force on a current carrying conductor is given by F = iℓB sin θ.
2. Here two sides PQ and RS of the coil are parallel to the field i.e. θ = 0 (sin θ = 0). Hence, the force on these two sides is zero.
3. The other two sides PS and QR make an angle of 90º with the field. Hence the force on each of these sides will be given by F = iLB.
4. Given i = 2.0 A, L = 25 cm = .25 m and B = 2.0 × 10⁻² W/m². F = 2.0 × 0.25 × (2.0 × 10⁻²) = 1.0 × 10⁻² N.
5. According to the Fleming's left-hand rule, these forces acting on the sides PS and QR are equal, parallel and opposite. Hence they constitute a couple whose moment = force F × perpendicular PS.
6. = (1.0 × 10⁻² N) (0.40 m) = 4.0 × 10⁻³ N.m.
7. Hence, the correct option is (B).

Teacher's Note:
a) 200 G = 200 × 10⁻⁴ T = 2.0 × 10⁻² T.
b) The sides of the coil parallel to the field experience no force, and the other two sides form a couple.

FORCE BETWEEN TWO PARALLEL CONDUCTORS

(a) When i₁ , i₂ are parallel. B at a distance 'd' due to i₁
B₁ = \( \frac{\mu_0 i_1}{2\pi d} \)

[Figure: Two long parallel wires X and Y a distance d apart carrying currents i₁ and i₂ in the same direction (upward in the figure); the field of X at Y is into the paper; the forces F₂₁ on X and F₁₂ on Y are shown pointing towards each other]

F₂₁ = Force on 2 due to 1
= i₂ ℓ B₁ , F₂₁ = \( \frac{\mu_0 i_1 i_2 \ell}{2\pi d} \)
force per unit length
= \( \frac{F_{21}}{\ell} \) = \( \frac{\mu_0 i_1 i_2}{2\pi d} \)
By a careful observation, we can say that |F₁₂| = |F₂₁| and F₁₂ = - F₂₁
So, the wires experience equal and opposite forces or there is an attractive force acting between wires.
(b) When i₁ , i₂ are antiparallel

[Figure: Two long parallel wires X and Y a distance d apart carrying currents I₁ and I₂ in opposite directions; the forces F₁₂ and F₂₁ are shown pointing away from each other]

in this case
F⃗₁₂ = - F⃗₂₁
\( \frac{|F_{12}|}{\ell} \) = \( \frac{|F_{21}|}{\ell} \) = \( \frac{\mu i_1 i_2}{2\pi d} \)
This force will be repulsive ⇒ parallel currents attract while antiparallel currents repel each other.

[Figure: Two long parallel wires PQ and ST, 30 cm apart; the current in PQ is 10 A and in ST is 15 A, both flowing from left to right]

Ex.27 In the adjoining fig the two parallel wires PQ and ST are at 30 cm apart. The currents flowing in the wires are according to fig. The force acting over a length of 5m of the wires is-
(A) 5 × 10⁻⁴ N, (attraction)
(B) 1 × 10⁻⁴ N, (attraction)
(c) 5 × 10⁻⁴ N, (repulsion)
(D) 1 × 10⁻⁴ N, (repulsion)

Answer: (A) 5 × 10⁻⁴ N, (attraction)

1. When currents flow in two long, parallel wires in the same direction, the wires exert a force of attraction on each other. The magnitude of this force acting per meter length of the wires is given by F = \( \frac{\mu_0}{2\pi}\,\frac{i_1 i_2}{R} \) = 2 × 10⁻⁷ \( \frac{i_1 i_2}{R} \) N/m.
2. Here i₁ = 10 A, i₂ = 15 A, R = 30 cm = 0.3 m.
3. ∴ F = 2 × 10⁻⁷ \( \frac{10 \times 15}{0.3} \) = 1 × 10⁻⁴ N/m.
4. ∴ Force on 5 m length of the wire = 5 × (1 × 10⁻⁴) = 5 × 10⁻⁴ N (attraction).
5. Hence, the correct option is (A).

Teacher's Note:
a) The figure shows 10 A in PQ and 15 A in ST, both flowing from left to right, so the currents are parallel.
b) Parallel currents attract; antiparallel currents repel.

[Figure: A long straight wire carrying the current i₁ and a rectangular loop ABCD carrying the current i₂ near it, with the distances a, b and d marked; the figure is not available in the source]

Ex.28 In the adjoining fig the force on current carrying loop ABCD due to a long current carrying (i₁) wire will be- (if i₁ = 20 A, i₂ = 16 A, a = 15 cm, b = 6 cm and d = 4 cm)
(A) 1.44 × 10⁻⁴ N, towards current - carrying wire
(B) 1.44 × 10⁻⁴ N, away from current - carrying wire
(C) 2.40 × 10⁻⁴ N, towards current - carrying wire
(D) 2.40 × 10⁻⁴ N, away from current - carrying wire

Answer: (B) 1.44 × 10⁻⁴ N, away from current - carrying wire

1. The (repulsive) force on the side AB of the loop, due to current i₁ is F₁ = \( \frac{\mu_0}{2\pi}\,\frac{i_1 i_2 L}{R} \), where \( \frac{\mu_0}{2\pi} \) = 2 × 10⁻⁷ N/A², i₁ = 20 A, i₂ = 16 A, R = d = 4 cm = 0.04 m and L = a = 15 cm = 0.15 m.
2. ∴ F₁ = (2 × 10⁻⁷) × \( \frac{20 \times 16 \times 0.15}{0.04} \) = 2.40 × 10⁻⁴ N. This force will act away from the current-carrying wire.
3. Similarly, the (attractive) force on the side CD of the loop, due to current i₁ is (now R = a + b = 10 cm = 0.10 m) F₂ = (2 × 10⁻⁷) × \( \frac{20 \times 16 \times 0.15}{0.10} \) = 0.96 × 10⁻⁴ N. This force will act towards the current-carrying wire.
4. The forces on the sides AD and BC of the loop will be equal and acting downward and upward respectively. Hence they will neutralize each other.
5. ∴ net force on the loop = F₁ - F₂ = (2.40 - 0.96) × 10⁻⁴ = 1.44 × 10⁻⁴ N acting away from the current - carrying wire.
6. Hence, the correct option is (B).

Teacher's Note:
a) If the current in the loop becomes clockwise, the net force on the loop will remain same, but its direction will be towards the current - carrying wire.
b) Special Note : The net force on a closed current carrying conductor placed in a magnetic field is zero.

[Figure: Three parallel wires D, C and G carrying currents with the wire C between D and G; the figure is not available in the source]

Ex. 29 In the following fig. the force experienced by a 35 cm. length of wire C will be-
(A) 4 × 10⁻⁴ N towards right
(B) 4 × 10⁻⁴ N towards left
(C) 3 × 10⁻⁴ N towards right
(D) 3 × 10⁻⁴ N towards left.

Answer: (C) 3 × 10⁻⁴ N towards right

1. The wire C is repelled by D as well as by G.
2. Force per unit length: \( \frac{F}{\ell} \) = \( \frac{\mu_0}{2\pi}\,\frac{I_D I_C}{0.03} \) (towards right) - \( \frac{\mu_0}{2\pi}\,\frac{I_C I_G}{0.05} \) (towards left).
3. Taking I_D = 30 A, I_C = 10 A and I_G = 20 A: \( \frac{F}{0.25} \) = 2 × 10⁻⁷ \( \left[ \frac{30 \times 10}{0.03} - \frac{10 \times 20}{0.05} \right] \).
4. ⇒ F = 3 × 10⁻⁴ N towards right.
5. Hence, the correct option is (C).

Teacher's Note:
a) The printed solution uses a length of 0.25 m, although the question says 35 cm; with 0.25 m the result 3 × 10⁻⁴ N matches option (C), with 35 cm it would be 4.2 × 10⁻⁴ N, which is not an option.
b) The figure is not available in the source, so the data are taken from the printed solution.

[Figure: A long straight wire carrying I₁ and a rectangular loop with the side XY at distance a from the wire and width b, carrying I₂; the figure is not available in the source]

Ex. 30 In the following fig. the force per unit length experienced by side XY of the rectangular loop, net force on the long straight wire and torque on the rectangular loop will be respectively-
(A) \( \frac{\mu_0 I_1 I_2}{2\pi a} \), 0, \( \frac{\mu_0 I_1 I_2 b}{2\pi a(a+b)} \)
(B) \( \frac{\mu_0 I_1 I_2 b}{2\pi a(a+b)} \), 0, \( \frac{\mu_0 I_1 I_2}{2\pi a} \)
(C) \( \frac{\mu_0 I_1 I_2}{2\pi a} \), \( \frac{\mu_0 I_1 I_2 b}{2\pi a(a+b)} \), 0
(D) None of the above.

Answer: (C)

1. (i) The force on XY due to I₁ is repulsive, and the force per unit length is F/L = \( \frac{\mu_0 I_1 I_2}{2\pi a} \).
2. (ii) Assuming that the action - reaction law is valid, the force on the long straight wire is repulsive due to current inside XY and attractive due to the other side. Thus \( \frac{F}{L} \) = \( \frac{\mu_0 I_1 I_2}{2\pi a} \) - \( \frac{\mu_0 I_1 I_2}{2\pi(a+b)} \) = \( \frac{\mu_0 I_1 I_2 b}{2\pi a(a+b)} \) and is repulsive.
3. (iii) Force on XY is \( \left( \frac{\mu_0 I_1 I_2}{2\pi a} \right) \), repulsive and force on the top side is \( \mu_0 \left\{ \frac{I_1 I_2}{2\pi(a+b)} \right\} \) attractive. These forces act in the plane of the loop. Therefore torque = 0. (these forces however tries to compress the loop)
4. Hence, the correct option is (C).

Teacher's Note:
a) The figure is not available in the source; the solution describes a long wire carrying I₁ with the rectangular loop of current I₂ whose side XY is at distance a and whose far side is at a + b.
b) Torque is zero because all the forces lie in the plane of the loop.

CURRENT CARRYING COIL IN AN EXTERNAL MAGNETIC FIELD

(a) As stated earlier, net force on a closed coil in a magnetic field is zero always but there acts a torque on it.
(b) torque on the coil is given by-
τ⃗ = M⃗ × B⃗
where M⃗ = Magnetic moment
B⃗ = Magnetic field
M⃗ = ni A⃗
(c) τ⃗ = Ni (A⃗ × B⃗) = BiNA sin θ.
Here A⃗ is an area vector whose direction is taken perpendicular to the plane.
The angle θ is angle between A⃗ and B⃗ (Note that it is not the angle between plane and B⃗)
(d) Work done to rotate a coil from an angle θ₁ to θ₂
W = \( \int_{\theta_1}^{\theta_2} \tau\, d\theta \) = \( \int_{\theta_1}^{\theta_2} BiNA \sin\theta\, d\theta \) = BiNA (cos θ₁ - cos θ₂)

Definition of B :

Force on a moving charge is given by
|F| = qvB sin θ
if θ = 90º, q = 1 coulomb, v = 1 m/s, then |F| = B
i.e. B is the force experienced by a unit charge moving with a unit velocity unit of
B = N/coulomb - m/sec = N-sec/(coulomb-m) = N/amp-m.

Ex.31 A 5 cm × 12 cm coil with number of turns 600 is placed in a magnetic field of strength 0.10 Tesla. The maximum magnetic torque acting on it when a current of 10⁻⁵ A is passed through it, will be
(A) 3.6 × 10⁻⁶ N-m
(B) 3.6 × 10⁻⁶ dyne-cm
(C) 3.6 × 10⁶ N-m
(D) 3.6 × 10⁶ dyne-m

Answer: (A) 3.6 × 10⁻⁶ N-m

1. τ_max = MB = ni AB = ni (ℓ × b) B.
2. τ_max = 600 × 10⁻⁵ × 5 × 10⁻² × 12 × 10⁻² × 0.10 = 3.6 × 10⁻⁶ N-m.
3. Hence, the correct answer is (A).

Teacher's Note:
a) The maximum torque occurs when the plane of the coil is parallel to the field.
b) Check: 600 × 10⁻⁵ = 6 × 10⁻³, area = 6 × 10⁻³ m², so 6 × 10⁻³ × 6 × 10⁻³ × 0.1 = 3.6 × 10⁻⁶ N-m.

Ex.32 The effective radius of a coil of 100 turns is 0.05 m and a current of 0.1 amp is flowing in it. The work required to turn this coil in an external magnetic field of 1.5 Tesla through 180º will be, if initially the plane of the coil is normal to the magnetic field.
(A) 0.236 Joule      (B) 0.236 Erg
(C) 236 Joule       (D) 236 erg.

Answer: (A) 0.236 Joule

1. The coil turns from θ = 0 to θ = 180º, so W = 2MB = 2πi Na²B.
2. W = 3.14 × 2 × 0.1 × 100 × (0.05)² × 1.5 = 0.236 Joule.
3. Hence, the correct answer is (A).

Teacher's Note:
a) W = MB (cos 0° - cos 180°) = 2MB.
b) The work needed is the difference between the potential energies at 180° and at 0°.

Ex.33 A conducting wire of length ℓ is turned in the form of a circular coil and a current i is passed through it. For torque due to magnetic field produced at its centre, to be maximum, the number of turns in the coil will be -
(A) 1            (B) 2
(C) any value   (D) more than 2

Answer: (A) 1

1. τ_max = MB or τ_max = niπa²B.
2. Let the number of turns in length ℓ be n, so ℓ = n(2πa) and a = \( \frac{\ell}{2\pi n} \).
3. τ_max = \( \frac{n i \pi B \ell^2}{4\pi^2 n^2} \) = \( \frac{\ell^2 i B}{4\pi n_{min}} \).
4. ∴ τ_max ∝ \( \frac{1}{n_{min}} \), and n_min = 1.
5. Hence, the correct option is (A).

Teacher's Note:
a) The question is printed as "(D) more than 2"; the answer is unaffected.
b) For a fixed length of wire, fewer turns mean a larger area, and area is squared while the turns count only once.

Ex.34 A circular coil of 20 turns and radius 10cm is placed in uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in the coil is 5A, then the torque acting on the coil will be -
(A) 31.4N-m
(B) 3.14 N-m
(C) 0.314 N-m
(D) zero

Answer: (D) zero

1. Torque acting on the coil is τ = NBIA sin θ.
2. The field is normal to the plane of the coil, so the angle between the area vector and the field is θ = 0º.
3. ∴ τ = 0.
4. Hence, the correct answer is (D).

Teacher's Note:
a) A torque exists only when the plane of the coil is not perpendicular to the field.
b) Check: sin 0° = 0, so no torque acts however large the current is.

POINTS TO REMEMBER

1. Lorentz Force zero If :
(a) The particle is uncharged (i.e. q = 0 ∵ F = Bq v sin θ = 0)
(b) The charged particle is at rest (i.e. v = 0)
(c) The charged particle is moving parallel or antiparallel to the field. (i.e. θ = 0 or 180º)
2. Lorentz force is maximum when the charged particle moves at right angles to the field.
3. Lorentz force F⃗ is perpendicular to the velocity vector v⃗ (i.e. F⃗ ⊥ v⃗ )
4. Lorentz force F⃗ is perpendicular to the field direction (i.e. F⃗ ⊥ B⃗)
5. Since, the Lorentz force is perpendicular to the direction of motion as well as to the field, it does not perform any work on the charged particle, therefore the magnitude of velocity of charged particle does not change, only the direction of motion changes.
6. The kinetic energy of the charged particle moving at right-angle to the field remains constant.
7. Total Lorentz force acting on a charged particle moving in electric and magnetic field is F⃗ = q ( E⃗ + v⃗ × B⃗ ).
8. Lorentz force is important in detecting the presence of electric and magnetic fields.
9. Lorentz force between two moving charges is F_m = \( \frac{\mu_0 q_1 q_2 v_1 v_2}{4\pi r^2} \)
10. Biot-Savart law for moving charge is B = \( \frac{\mu_0 q v \sin\theta}{4\pi r^2} \)
11. If the charges move with speeds comparable to the speed of light, magnetic and electric forces between them would become comparable.
12. Magnetic field due to a toroid B ∝ n and B ∝ a
Where a → radius of toroid and n → number of turns per unit length.
13. The magnetic dipole moment is a vector quantity, directed along the axis of the dipole. The direction of the magnetic moment of a current-loop is determined by the right-hand rule.
14. Right - hand rule : "If we stretch our right-hand palm and curl the fingers around the loop in the direction of the current, then the thumb will point in the direction of the magnetic moment".
15. A magnetic dipole is in stable equilibrium if M⃗ and B⃗ are parallel to each other, while the magnetic dipole is in unstable equilibrium if M⃗ and B⃗ are antiparallel to each other.
16. Repulsion is the surest test for distinguishing between a magnet and a piece of iron : The pole of a magnet attracts a piece of iron as well as the unlike pole of an other magnet. Therefore, when a testing magnet is brought near the two ends of a specimen, it will show attraction whether it is an iron rod or a magnet. Thus, repulsion is the surest test to distinguish between a magnet and a piece of iron.
17. Most of the magnetic moment is produced due to the electron spin : the contribution of the orbital revolution is very small.
18. It has been observed experimentally that the positions of the earth's magnetic poles change gradually.