CBSE Class 12 Physics Magnetic Effects Of Current And Magnetism chapter notes and important questions

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MAGNETIC EFFECTS OF CURRENT AND MAGNETISM

Magnetic Effects of Current and Magnetism – gist of the chapter followed by questions with answers

GIST OF THE CHAPTER

1. Magnetic field: the region around a magnet or a current-carrying conductor in which its magnetic influence can be felt by a magnetic needle.
2. Biot–Savart law: \(dB=\dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin\theta}{r^2}\), with \(\mu_0=4\pi\times10^{-7}\) T m/A. The direction of \(dB\) is given by Maxwell’s right-hand (corkscrew) rule.
3. Applications of the Biot–Savart law: field at the centre of a circular coil, \(B=\dfrac{\mu_0I}{2a}\) (for \(N\) turns \(\dfrac{\mu_0NI}{2a}\)); field on the axis of a circular coil, \(B=\dfrac{\mu_0NIa^2}{2(a^2+x^2)^{3/2}}\), where \(N\) is the number of turns.
4. Ampere’s circuital law: the line integral of the magnetic field around any closed path in vacuum/air is \(\mu_0\) times the total current threading the closed path, \(\oint\vec B\cdot d\vec l=\mu_0I\).
5. Applications of Ampere’s law: (i) infinitely long straight conductor, \(B=\dfrac{\mu_0I}{2\pi r}\); (ii) straight solenoid, \(B=\mu_0nI\), where \(n\) is the number of turns per unit length; (iii) toroidal solenoid, \(B=\dfrac{\mu_0NI}{2\pi r}\), where \(N\) is the total number of turns.
6. Force on a moving charge (Lorentz force): in a magnetic field \(\vec F=q(\vec v\times\vec B)\); in electric and magnetic fields \(\vec F=q[\vec E+(\vec v\times\vec B)]\).
7. Cyclotron: Principle: (a) a charged particle moving at right angles to a uniform magnetic field describes a circular path; (b) an ion can acquire a sufficiently large energy with a low ac voltage by crossing the same electric field repeatedly under a strong magnetic field. Cyclotron (magnetic resonance) frequency \(\nu=\dfrac{qB}{2\pi m}\), \(T=\dfrac{2\pi m}{Bq}\), \(\omega=\dfrac{Bq}{m}\). Maximum velocity \(v_m=\dfrac{Bqr_m}{m}\) and maximum kinetic energy \(E_m=\dfrac{B^2q^2r_m^2}{2m}\).
8. Force on a current-carrying conductor in a uniform field: \(\vec F=I(\vec l\times\vec B)\), where \(l\) is the length of the conductor. Its direction is given by Fleming’s left-hand rule.
9. Force per unit length between parallel infinitely long current-carrying conductors: \(\dfrac Fl=\dfrac{\mu_0I_1I_2}{2\pi d}\). (a) Currents in the same direction: the wires attract each other. (b) Currents in opposite directions: they repel each other.
10. One ampere is that current which, when flowing through each of two parallel straight conductors of infinite length placed in free space 1 m apart, produces between them a force of \(2\times10^{-7}\) N per metre of their length.
11. Torque on a current loop in a uniform field: \(\tau=NIBA\sin\theta\), i.e. \(\vec\tau=\vec M\times\vec B\), where \(M=NIA\).
12. Motion of a charged particle in a magnetic field: (a) perpendicular to the field, \(F=qvB\sin90^\circ=qvB\) (circular path); (b) parallel or antiparallel to the field, \(F=qvB\sin0^\circ=0\) (straight-line path); (c) if \(0\lt\theta\lt90^\circ\) the path is a helix: \(v\cos\theta\) is responsible for the linear motion and \(v\sin\theta\) for the circular motion.
13. Moving coil galvanometer: a sensitive instrument used for detecting small electric currents. Principle: a current-carrying coil placed in a magnetic field experiences a torque. \(NIAB=k\theta\), so \(I=G\theta\) where \(G=\dfrac{k}{NBA}\) is the galvanometer constant (\(k\) = torsional constant of the spring). Current sensitivity \(I_s=\dfrac\theta I=\dfrac{NBA}{k}\); voltage sensitivity \(V_s=\dfrac\theta V=\dfrac{NBA}{kR}\).
14. Galvanometer to ammeter: a small resistance \(S\) is connected in parallel with the galvanometer coil, \(S=\dfrac{I_gG}{I-I_g}\); \(R_A=\dfrac{GS}{G+S}\). Galvanometer to voltmeter: a high resistance \(R\) is connected in series, \(R=\dfrac V{I_g}-G\); \(R_V=G+R\).
15. Current loop as a magnetic dipole: for an electron in a circular orbit, \(M=\dfrac{evr}{2}\); in terms of the Bohr magneton, \(M=n\dfrac{eh}{4\pi m_e}\).
16. Uniform magnetic field: represented by equally spaced parallel lines; crosses (×) represent a field directed into the page.
17. Magnetic dipole moment of a bar magnet: \(M=m(2l)\), where \(m\) is the pole strength; the SI unit of \(M\) is A m².
18. Work done in rotating a magnetic dipole from \(\theta_1\) to \(\theta_2\) in a field: \(W=MB(\cos\theta_1-\cos\theta_2)\). The potential energy is \(U=-\vec M\cdot\vec B\). Torque on a dipole: \(\vec\tau=\vec M\times\vec B\).
19. Field due to a magnetic dipole: (a) on the axial line, \(B=\dfrac{\mu_0}{4\pi}\dfrac{2Mr}{(r^2-l^2)^2}\); (b) on the equatorial line, \(B=\dfrac{\mu_0}{4\pi}\dfrac{M}{(r^2+l^2)^{3/2}}\). For a short dipole these become \(\dfrac{\mu_0}{4\pi}\dfrac{2M}{r^3}\) and \(\dfrac{\mu_0}{4\pi}\dfrac{M}{r^3}\).
20. Elements of the earth’s magnetic field: (a) Declination (\(\theta\)): the angle between the magnetic meridian and the geographic meridian. (b) Inclination or dip (\(\delta\)): the angle between the direction of the total intensity of the earth’s field and its horizontal component. (c) Horizontal component \(B_H\): the component of the total intensity along the horizontal. \(\tan\delta=\dfrac{B_V}{B_H}\) and \(B=\sqrt{B_V^2+B_H^2}\).
21. Magnetic quantities: (a) intensity of magnetisation \(I=\dfrac MV=\dfrac{\text{magnetic moment}}{\text{volume}}\), with \(M=m\cdot2l\); (b) magnetic induction \(B=\mu_0(H+I)\), where \(H\) is the magnetising field; (c) permeability \(\mu=\dfrac BH\) (ratio of magnetic induction to magnetic intensity); (d) susceptibility \(\chi=\dfrac IH\).
22. Electromagnet and permanent magnet: (a) an electromagnet is a magnet whose magnetism is due to current flowing through a coil wound over soft iron; it remains magnetised only as long as the current flows in the coil. The core material should have low retentivity and low coercivity (and high permeability), e.g. soft iron. (b) A permanent magnet owes its strength to the alignment of its molecules, e.g. steel; the material should have high retentivity and high coercivity.

Properties of magnetic substances

Rows 4 and 5 of the table (figures): a freely suspended rod between the magnetic poles (diamagnetic: perpendicular to the field; paramagnetic and ferromagnetic: parallel to the field), and a liquid in a watch glass between the poles (closer poles, upper; farther poles, lower).

Graph between \(H\) and \(I\): for a ferromagnet (steep, large positive), a paramagnet (small positive) and a diamagnet (negative). Graph between \(\chi_m\) and temperature \(T\): constant for a diamagnet, decreasing with \(T\) (\(\chi_m\propto1/T\)) for a paramagnet, and decreasing with \(T\) up to the Curie point for a ferromagnet.

QUESTIONS WITH ANSWERS

In a certain arrangement, a proton does not get deflected while passing through a magnetic field region. State the condition under which it is possible.

Answer: The velocity \(\vec v\) of the proton is parallel or antiparallel to \(\vec B\).

An electron beam is moving vertically upwards. If it passes through a magnetic field directed from South to North in a horizontal plane, in what direction will the beam be deflected?

Answer: Towards the geographical East in the horizontal plane.

What is the work done by the magnetic force on a charged particle moving perpendicular to the magnetic field?

Answer: Zero.

A current is set up in a long copper pipe. What is the magnetic field inside the pipe?

Answer: Zero (no current is enclosed by a loop drawn inside the pipe; Ampere’s law).

How does the value of the angle of dip vary on moving from the equator to the poles?

Answer: It varies from 0° to 90°.

A compass needle in a horizontal plane is taken to geographic north / south poles. In what direction does the needle align?

Answer: At the poles the horizontal component of the earth’s field is (nearly) zero, so the needle has no preferred direction and can rest in any direction.

The horizontal component of the earth’s magnetic field is 0.2 G and the total magnetic field is 0.4 G. Find the angle of dip.

Answer: \(\cos\delta=\dfrac{B_H}{B}=\dfrac{0.2}{0.4}=\dfrac12\), so \(\delta=60^\circ\).

An iron bar magnet is heated to 1000 °C and then cooled in a magnetic field free space. Will it retain its magnetism?

Answer: No. 1000 °C is above the Curie temperature of iron (about 770 °C); the thermal motion destroys the alignment of the domains, and on cooling in a field-free space they remain randomly oriented, so the bar is not magnetised.

What is the net magnetic moment of an atom of a diamagnetic material?

Answer: Zero.

Which materials have a negative value of magnetic susceptibility?

Answer: Diamagnetic materials.

Why are permanent magnets made of steel while the core of a transformer is made of soft iron?

Answer: Steel has high retentivity and high coercivity, so it keeps its magnetism and is not easily demagnetised. Soft iron has low coercivity and high permeability, so it is easily magnetised and demagnetised in every ac cycle with very little energy loss (small hysteresis loop).

Why is a radial magnetic field used in a galvanometer?

Answer: A radial field makes the deflection directly proportional to the current, which gives a linear current scale.

A particle of mass ‘m’ and charge ‘q’ moving with velocity ‘v’ enters a region of uniform magnetic field at right angle to the direction of its motion. How does its kinetic energy get affected?

Answer: The kinetic energy is not affected.

What can be the cause of helical motion of a charged particle?

Answer: The charged particle moves inclined to the magnetic field.

An electron is moving with velocity \(v\) along the axis of a long straight solenoid carrying current \(I\). What will be the force acting on the electron due to the magnetic field of the solenoid?

Answer: Zero, because the field inside the solenoid is along the axis, so the velocity is parallel to \(B\) and \(F=qvB\sin0^\circ=0\).

Write the condition under which an electron does not experience a force in a magnetic field.

Answer: (i) When the electron is stationary, or (ii) when \(\vec v\parallel\vec B\).

Write the S.I. units of (i) magnetic pole strength (ii) magnetic dipole moment.

Answer: (i) A m (ii) A m².

Why should the material used for making permanent magnets have high coercivity?

Answer: So that the magnetisation is not easily erased by stray magnetic fields, minor mechanical damage or temperature variation.

The permeability of a magnetic material is 0.9987. Name the type of material it represents.

Answer: Diamagnetic, because \(\mu\lt1\) (relative permeability less than 1).

How are the magnetic field lines different from the electrostatic field lines?

Answer: Magnetic field lines form closed loops, while electric field lines originate from positive charges and end at negative charges.

Mention two properties of soft iron due to which it is preferred for making an electromagnet.

Answer: Low coercivity and high permeability.

How does the magnetic moment of an electron in a circular orbit of radius \(r\) and moving with a speed \(v\) change when the frequency of revolution is doubled?

Answer: \(M=IA=efA\); when \(f'=2f\), \(M'=2M\) (it doubles).

A given rectangular coil OLMN of area \(A\), carrying a given current \(I\), is placed in a uniform magnetic field \(\vec B=B\hat k\) in two different orientations (a) and (b). What is the magnitude of the torque experienced by this coil in the two cases?

Answer: (a) The plane of the coil is perpendicular to the field’s direction of the moment: \(\tau=mB\sin90^\circ=IAB\). (b) \(\tau=mB\sin0^\circ=0\). (The two orientations are shown in the original figure: in (a) the magnetic moment of the coil is perpendicular to \(B\), in (b) it is parallel to \(B\).)

Two circular loops, of radii \(r\) and \(2r\), have currents \(I\) and \(I/2\) flowing through them in clockwise and anticlockwise sense respectively. If their magnetic moments are \(\vec M_1\) and \(\vec M_2\), state the relation between them.

Answer: \(\vec M_1=-\pi r^2I\hat k\) and \(\vec M_2=\pi(2r)^2\dfrac I2\hat k=2\pi r^2I\hat k\). Hence \(\vec M_2=-2\vec M_1\) (\(M_2\) is twice \(M_1\) and opposite in direction).

Which criterion is used to select a suitable ferromagnet for transformer cores?

Answer: The ferromagnet must have a small area of the hysteresis loop, to minimise the energy loss in cycles of magnetisation and demagnetisation.

In what way is the behaviour of a diamagnetic material different from that of a paramagnetic material when kept in an external magnetic field?

Answer: A diamagnetic specimen moves towards the weaker region of the field, while a paramagnetic specimen moves towards the stronger region. A diamagnetic specimen is repelled by a magnet, while a paramagnetic specimen moves towards the magnet. A paramagnetic rod gets aligned along the field and a diamagnetic rod perpendicular to the field.

In what direction will a compass needle point when kept at the (i) poles and (ii) equator?

Answer: At the poles the compass needle can point in any direction; at the equator it is parallel to the earth’s surface.

The vertical component of the earth’s magnetic field at a place is \(\sqrt3\) times the horizontal component. What is the value of the angle of dip?

Answer: \(\tan\delta=\dfrac{B_V}{B_H}=\sqrt3\Rightarrow\delta=60^\circ\).

What is the value of the horizontal component of the earth’s magnetic field at the magnetic poles?

Answer: \(B_H=0\).

What should be the orientation of a magnetic dipole in a uniform magnetic field so that its potential energy is maximum?

Answer: When \(\vec M\) and \(\vec B\) are antiparallel: \(U=-MB\cos180^\circ=+MB\), the maximum value.

Explain how the Biot–Savart law enables one to express Ampere’s circuital law in the integral form, viz. \(\oint\vec B\cdot d\vec l=\mu_0I\).

Answer: The Biot–Savart law can be expressed as Ampere’s circuital law by considering the surface to be made up of a large number of loops. The sum of the tangential components of the magnetic field multiplied by the length of all such elements gives the result \(\oint\vec B\cdot d\vec l=\mu_0I\).

A wire of length 0.04 m carrying a current of 12 A is placed inside a solenoid, making an angle of 30° with its axis. The field due to the solenoid is 0.25 T. Find the force on the wire.

Answer: \(F=BIl\sin\theta=0.25\times12\times0.04\times\sin30^\circ=0.06\) N.

A circular loop of radius 0.1 m carries a current of 1 A and is placed in a uniform magnetic field of 0.5 T. The magnetic field is perpendicular to the plane of the loop. What is the force experienced by the loop?

Answer: A magnetic dipole does not experience any net force in a uniform magnetic field. Hence the current-carrying loop does not experience any net force.

A proton, an alpha particle and a deuteron are moving in circular paths with the same kinetic energies in the same magnetic field. Find the ratio of their radii and time periods.

Answer: \(r=\dfrac{\sqrt{2mK}}{qB}\propto\dfrac{\sqrt m}{q}\): \(r_p:r_\alpha:r_d=1:1:\sqrt2\). \(T=\dfrac{2\pi m}{qB}\propto\dfrac mq\): \(T_p:T_\alpha:T_d=1:2:2\).

An electron moving with kinetic energy 25 keV moves perpendicular to a uniform magnetic field of 0.2 mT. Calculate the time period of rotation of the electron in the magnetic field.

Answer: \(T=\dfrac{2\pi m}{eB}=\dfrac{2\pi\times9.1\times10^{-31}}{1.6\times10^{-19}\times0.2\times10^{-3}}=1.79\times10^{-7}\) s (independent of the kinetic energy).

A wire placed along the north–south direction carries a current of 5 A from South to North. Find the magnetic field due to a 1 cm piece of wire at a point 200 cm North-East from the piece.

Answer: \(dB=\dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin\theta}{r^2}=10^{-7}\times\dfrac{5\times0.01\times\sin45^\circ}{(2)^2}\approx8.8\times10^{-10}\) T, acting vertically downwards.

How will the magnetic field intensity at the centre of a circular coil carrying current change if the current through the coil is doubled and the radius of the coil is halved?

Answer: \(B=\dfrac{\mu_0NI}{2r}\). With \(I\to2I\) and \(r\to r/2\): \(B'=\dfrac{\mu_0N(2I)}{2(r/2)}=4B\). The field becomes four times.

A circular coil of 500 turns has a radius of 2 m and carries a current of 2 A. What is the magnetic field at a point on the axis of the coil at a distance equal to the radius of the coil from the centre?

Answer: \(B=\dfrac{\mu_0NIR^2}{2(R^2+x^2)^{3/2}}\) with \(x=R\): \(B=\dfrac{\mu_0NI}{2\cdot2\sqrt2\,R}=\dfrac{4\pi\times10^{-7}\times500\times2}{4\sqrt2\times2}=1.11\times10^{-4}\) T.

The strength of magnetic induction at the centre of a current-carrying circular coil is \(B_1\) and at a point on its axis at a distance equal to its radius from the centre is \(B_2\). Find \(B_1/B_2\).

Answer: \(B_1=\dfrac{\mu_0NI}{2R}\), \(B_2=\dfrac{\mu_0NIR^2}{2(2R^2)^{3/2}}=\dfrac{\mu_0NI}{4\sqrt2R}\). So \(\dfrac{B_1}{B_2}=2\sqrt2\).

A current is flowing in a circular coil of radius ‘r’ and the magnetic field at its centre is \(B_0\). At what distance from the centre on the axis of the coil will the magnetic field be \(B_0/8\)?

Answer: \(\dfrac{B}{B_0}=\dfrac{r^3}{(r^2+x^2)^{3/2}}=\dfrac18\Rightarrow r^2+x^2=4r^2\Rightarrow x=\sqrt3\,r\).

A long straight solid metal wire of radius ‘R’ carries a current ‘I’, uniformly distributed over its circular cross-section. Find the magnetic field at a distance ‘r’ from the axis of the wire (a) inside and (b) outside the wire.

Answer: (a) Inside (\(r\lt R\)): \(B=\dfrac{\mu_0Ir}{2\pi R^2}\) (for a material of relative permeability \(\mu_r\), multiply by \(\mu_r\)). (b) Outside (\(r\gt R\)): \(B=\dfrac{\mu_0I}{2\pi r}\).

Find the value of the magnetic field inside a hollow straight current-carrying conductor at a distance \(r\) from the axis of the conductor.

Answer: \(B=0\) (no current is enclosed by an Amperian loop drawn inside the hollow).

A rectangular loop of size 25 cm × 10 cm carrying a current of 15 A is placed 2 cm away from a long straight conductor carrying a current of 25 A. What is the direction and magnitude of the net force acting on the loop?

Answer: The 25 cm sides are parallel to the wire. The side nearer the wire (2 cm away) carries current in the same direction as the wire and is attracted; the far side (12 cm away) is repelled with a smaller force; the forces on the 10 cm sides cancel.

\[F=\dfrac{\mu_0I_1I_2l}{2\pi}\left[\dfrac1{0.02}-\dfrac1{0.12}\right]=2\times10^{-7}\times25\times15\times0.25\times41.67=7.8\times10^{-4}\ \text{N}\]

The net force is about \(7.8\times10^{-4}\) N, directed towards the long wire.

A long straight conductor PQ, carrying a current of 60 A, is fixed horizontally. Another long conductor XY is kept parallel to PQ at a distance of 4 mm, in air. Conductor XY is free to move and carries a current ‘I’. Calculate the magnitude and direction of the current ‘I’ for which the magnetic repulsion just balances the weight of the conductor XY.

Answer: The magnetic force per unit length \(\dfrac{\mu_0\,60\,I}{2\pi(4\times10^{-3})}\) must equal the weight per unit length \(\lambda g\). The mass per unit length of XY is not stated in the question; the given answer corresponds to \(\lambda=0.01\) kg/m (\(g=9.8\) m/s²): \(3\times10^{-3}\,I=0.098\), so \(I=32.67\) A.

The current in XY must flow opposite to that in PQ, because only then is the force repulsive.

A circular coil of 200 turns and radius 5 cm carries a current of 2.5 A. It is suspended vertically in a uniform horizontal magnetic field of 0.25 T, with the plane of the coil making an angle of 60° with the field lines. Calculate the magnitude of the torque that must be applied on it to prevent it from turning.

Answer: The angle between the normal (magnetic moment) and the field is \(90^\circ-60^\circ=30^\circ\). \(\tau=NIAB\sin30^\circ=200\times2.5\times\pi(0.05)^2\times0.25\times0.5=0.49\) N m.

A short bar magnet has a magnetic moment of 50 A m². Calculate the magnetic field intensity at a distance of 0.2 m from its centre on (1) its axial line (2) its equatorial line.

Answer: (1) \(B_1=\dfrac{\mu_0}{4\pi}\dfrac{2M}{r^3}=10^{-7}\times\dfrac{100}{0.008}=1.25\times10^{-3}\) T. (2) \(B_2=\dfrac{\mu_0}{4\pi}\dfrac{M}{r^3}=0.625\times10^{-3}\) T.

Calculate the torque acting on a magnet of length 20 cm and pole strength \(2\times10^{-5}\) A m, placed in the earth’s magnetic field of flux density \(2\times10^{-5}\) T, when (a) the magnet is parallel to the field (b) the magnet is perpendicular to the field.

Answer: \(M=m\times2l=2\times10^{-5}\times0.2=4\times10^{-6}\) A m². (a) \(\tau=MB\sin0^\circ=0\). (b) \(\tau=MB=4\times10^{-6}\times2\times10^{-5}=0.8\times10^{-10}\) N m.

A long straight horizontal cable carries a current of 2.5 A in the direction 10° south of west to 10° north of east. The magnetic meridian of the place happens to be 10° west of the geographic meridian. The earth’s magnetic field at the location is 0.33 G and the angle of dip is zero. Ignoring the thickness of the cable, locate the line of neutral points.

Answer: The cable is perpendicular to the magnetic meridian. At a neutral point the field of the cable equals the horizontal component of the earth’s field and is opposite to it: \(B_H=\dfrac{\mu_0I}{2\pi r}\), so \(r=\dfrac{2\times10^{-7}\times2.5}{0.33\times10^{-4}}=1.5\) cm. The neutral points lie on a line parallel to the cable, 1.5 cm vertically above it.

The vertical component of the earth’s magnetic field at a place is \(\sqrt3\) times the horizontal component. What is the value of the angle of dip at this place?

Answer: \(\tan\delta=\sqrt3\), so \(\delta=60^\circ\).

A ship is sailing due west according to a mariner’s compass. If the declination of the place is 15° east, what is the true direction of the ship?

Answer: Magnetic north is 15° east of true north, so the magnetic west direction is rotated by 15° the same way. The true direction is 75° west of north.

The susceptibility of a magnetic material is 0.9853. Identify the type of the magnetic material. Draw the modification of the field pattern on keeping a piece of this material in a uniform magnetic field.

Answer: The susceptibility is positive, so the material is paramagnetic (the field lines crowd into the specimen).

A long wire is first bent into a circular coil of one turn and then into a circular coil of smaller radius having \(n\) turns. If the same current passes in both the cases, find the ratio of the magnetic fields produced at the centres in the two cases.

Answer: One turn of radius \(r\): \(B=\dfrac{\mu_0I}{2r}\). For \(n\) turns of radius \(r'\): \(n(2\pi r')=2\pi r\), so \(r'=r/n\). Then \(B'=\dfrac{\mu_0nI}{2r'}=\dfrac{\mu_0n^2I}{2r}=n^2B\). The ratio \(B:B'=1:n^2\).

A beam of α-particles and of protons, of the same velocity \(v\), enters a uniform magnetic field at right angles to the field lines. The particles describe circular paths. Calculate the ratio of the radii of the two paths.

Answer: \(Bqv=\dfrac{mv^2}{r}\), so \(r=\dfrac{mv}{Bq}\). \(\dfrac{r_\alpha}{r_p}=\dfrac{m_\alpha/q_\alpha}{m_p/q_p}=\dfrac{4/2}{1/1}=2\).

To increase the current sensitivity of a moving coil galvanometer by 50% its resistance is increased so that the new resistance becomes twice its initial resistance. By what factor does its voltage sensitivity change?

Answer: \(V_s=\dfrac{I_s}{R}\). New \(I_s'=1.5I_s\) and \(R'=2R\): \(V_s'=\dfrac{1.5I_s}{2R}=0.75V_s\). The voltage sensitivity becomes 0.75 times (decreases by 25%).

Find the magnitude of the force on each segment of the wire ABCDE, if a magnetic field of 0.30 T is applied parallel to AB and DE. Take the value of the current flowing in the wire as 1 A. (AB = 5 cm, BC = 8 cm, CD = 10 cm, DE = 10 cm.)

Answer: \(F=IlB\sin\theta\). Segments AB and DE are parallel to the field, so the force on them is zero. BC is perpendicular to the field: \(F_{BC}=1\times0.08\times0.30\times\sin90^\circ=0.024\) N. CD makes 30° with the field: \(F_{CD}=1\times0.10\times0.30\times\sin30^\circ=0.015\) N.

Write the relation for current sensitivity and voltage sensitivity of a moving coil galvanometer. Using these relations, explain the fact that increasing the current sensitivity may not necessarily increase the voltage sensitivity.

Answer: Current sensitivity (deflection per unit current): \(\dfrac\phi I=\dfrac{NAB}{k}\). Voltage sensitivity (deflection per unit voltage): \(\dfrac\phi V=\dfrac{NAB}{R_gk}\).

If we double the number of turns of the coil, the current sensitivity doubles. But the resistance of the galvanometer also doubles, since it is proportional to the length of the wire. Thus the voltage sensitivity remains unchanged.

Why do two wires carrying current in the same direction attract each other?

Answer: Each current-carrying conductor produces a magnetic field around it, so each conductor experiences a force due to the magnetic field of the other. By Fleming’s left-hand rule the force on each wire is perpendicular to the wires and directed towards the other wire. Hence the two wires attract each other.

A galvanometer coil of 50 Ω resistance shows full scale deflection for a current of 5 mA. How will you convert this galvanometer into a voltmeter of range 0 to 15 V?

Answer: The galvanometer is converted into a voltmeter by connecting a resistance \(R\) in series with its coil: \(R=\dfrac V{I_g}-R_g=\dfrac{15}{5\times10^{-3}}-50=2950\ \Omega\).

An element of length 1 cm along the x-axis is placed at the origin and carries a large current \(I=10\) A in the positive x-direction. What is the magnitude and direction of the magnetic field on the positive y-axis at 0.5 m from the origin?

Answer: \(d\vec B=\dfrac{\mu_0}{4\pi}\dfrac{I\,d\vec l\times\hat r}{r^2}=10^{-7}\times\dfrac{10\times0.01}{0.25}(\hat i\times\hat j)=4\times10^{-8}\ \hat k\) T. The magnitude is \(4\times10^{-8}\) T, directed along the positive z-axis.

A current ‘I’ enters a uniform circular loop of radius ‘R’ at point M and flows out at N, the angle MON being 90°. Obtain the net magnetic field at the centre of the loop.

Answer: The current divides into \(I_1\) in the longer arc (270°) and \(I_2\) in the shorter arc (90°). The resistance of an arc is proportional to its length, so the longer arc has three times the resistance and \(I_2=3I_1\). The field at the centre due to an arc is \(B=\dfrac{\mu_0I}{4\pi R}\theta\). So \(B_2=\dfrac{\mu_0I_2}{8R}\) (shorter arc) and \(B_1=\dfrac{3\mu_0I_1}{8R}\) (longer arc), in opposite directions (one into, the other out of the page). Since \(I_2=3I_1\), \(B_2=B_1\) and the net field \(B=0\).

A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires. Find the magnitude and direction of the magnetic field which should be set up so that the tension in the wires is zero.

Answer: For zero tension the magnetic force must balance the weight: \(IlB=mg\), so \(B=\dfrac{mg}{Il}=\dfrac{60\times10^{-3}\times9.8}{5\times0.45}=0.26\) T. The magnetic field must be directed vertically inwards (into the page, as in the figure) so that the force on the rod is upward.

Two long coaxial insulated solenoids \(S_1\) and \(S_2\) of equal lengths are wound one over the other. A steady current \(I\) flows through the inner solenoid \(S_1\) to the other end B, which is connected to the outer solenoid \(S_2\) through which the same current \(I\) flows in the opposite direction so as to come out at end A. If \(n_1\) and \(n_2\) are the numbers of turns per unit length, find the magnitude and direction of the net magnetic field at a point (i) inside on the axis and (ii) outside the combined system.

Answer: The field on the axis of a solenoid is \(B=\mu_0nI\). (i) Inside the combined system on the axis: \(B=B_1-B_2=\mu_0n_1I-\mu_0n_2I=\mu_0I(n_1-n_2)\), in the direction of the field of \(S_1\) if \(n_1\gt n_2\). (ii) Outside the combined system the net field is zero.

Show diagrammatically the behaviour of magnetic field lines in the presence of (i) paramagnetic and (ii) diamagnetic substances. How does one explain this distinguishing feature?

Answer: A paramagnetic material tends to move from weaker to stronger regions of the magnetic field and hence increases the number of lines of the magnetic field passing through it. (Alternatively: in a paramagnetic material, dipole moments are induced in the direction of the field.)

A diamagnetic material tends to move from stronger to weaker regions of the magnetic field and hence decreases the number of lines of magnetic field passing through it. (Alternatively: in a diamagnetic material, dipole moments are induced in the opposite direction of the field.)

To which of the two – a polar dielectric or a non-polar dielectric – does a paramagnetic material correspond? Justify your answer.

Answer: A paramagnetic material corresponds to a polar dielectric. The atoms/molecules of a polar dielectric have a non-zero dipole moment, and the atoms/molecules of a paramagnetic material likewise have a non-zero magnetic moment (a non-polar dielectric has zero dipole moment).

A square-shaped plane coil of area 100 cm² of 200 turns carries a steady current of 5 A. It is placed in a uniform magnetic field of 0.2 T acting perpendicular to the plane of the coil. Calculate the torque on the coil when its plane makes an angle of 60° with the direction of the field. In which orientation will the coil be in stable equilibrium?

Answer: Magnetic moment \(m=NIA=200\times5\times100\times10^{-4}=10\) A m². The angle between the normal and the field is \(90^\circ-60^\circ=30^\circ\): \(\tau=mB\sin30^\circ=10\times0.2\times0.5=1\) N m. The coil is in stable equilibrium when its plane is perpendicular to the field, i.e. its magnetic moment is parallel to \(B\).

A diamagnetic material and a paramagnetic material of the same shape and size are in turn kept in an external uniform magnetic field. Draw the modification of the magnetic lines of force in the two cases. How does the intensity of magnetisation of a paramagnetic material vary with temperature?

Answer: The modification of the field lines is as shown above (paramagnetic: lines crowd into the material; diamagnetic: lines are pushed out). The intensity of magnetisation of a paramagnetic material is inversely proportional to its absolute temperature (Curie’s law).

(i) Name the three elements of the earth’s magnetic field. (ii) Where on the surface of the earth is the vertical component of the earth’s magnetic field zero?

Answer: (i) The angle of declination, the angle of inclination (dip) and the horizontal component of the earth’s magnetic field. (ii) At the equator.

Using the relation for the potential energy of a current-carrying planar loop in a uniform magnetic field, obtain the expression for the work done in moving the planar loop from its unstable equilibrium position to its stable equilibrium position.

Answer: \(U=-\vec M\cdot\vec B=-MB\cos\theta\), where \(\theta\) is the angle between \(\vec M\) and \(\vec B\). Unstable equilibrium: \(\theta=180^\circ\), \(U=+MB\). Stable equilibrium: \(\theta=0^\circ\), \(U=-MB\). The work done is \(W=U_{final}-U_{initial}=-MB-MB=-2MB\); that is, the field does work \(2MB\) in turning the loop to the stable position.

(a) If magnetic monopoles were to exist, how would Gauss’s law of magnetism get modified? (b) How will the angle of dip vary when one goes from a place where the acceleration due to gravity is maximum to a place where it is minimum, on the surface of the earth?

Answer: (a) \(\oint\vec B\cdot d\vec S=\mu_0q_m\), where \(q_m\) is the net pole strength enclosed by the surface. (b) The acceleration due to gravity is maximum at the poles and minimum at the equator, so the angle of dip varies from 90° (at the poles) to 0° (at the equator).

Two magnets of magnetic moments \(M\) and \(M\sqrt3\) are joined to form a cross. The combination is suspended in a uniform magnetic field \(B\). The magnetic moment \(M\) now makes an angle \(\theta\) with the field direction. Find the value of the angle \(\theta\).

Answer: The net torque on the combination is zero: \(MB\sin\theta-\sqrt3MB\sin(90^\circ-\theta)=0\Rightarrow\sin\theta=\sqrt3\cos\theta\Rightarrow\tan\theta=\sqrt3\Rightarrow\theta=60^\circ\).

What is the basic difference between the atom or molecule of a diamagnetic and a paramagnetic material? Why are elements with even atomic number more likely to be diamagnetic?

Answer: The atom/molecule of a diamagnetic material has zero net magnetic moment; for a paramagnetic material it is not so. With an even atomic number, the electrons in an atom can ‘pair off’, which makes the net magnetic moment of each pair zero. This makes the element more likely to be diamagnetic.

A magnetising field of 1500 A/m produces a flux of \(2.4\times10^{-5}\) weber in a bar of iron of cross-sectional area 0.5 cm². Calculate the permeability and susceptibility of the iron bar used.

Answer: \(B=\dfrac\phi A=\dfrac{2.4\times10^{-5}}{0.5\times10^{-4}}=0.48\) T. Permeability \(\mu=\dfrac BH=\dfrac{0.48}{1500}=3.2\times10^{-4}\) Wb/Am. \(\mu_r=\dfrac{\mu}{\mu_0}=\dfrac{3.2\times10^{-4}}{4\pi\times10^{-7}}\approx255\). Susceptibility \(\chi=\mu_r-1=254\).

A wire AB is carrying a current of 12 A and is lying on the table. Another wire CD, carrying a current of 5 A, is arranged just above AB at a height of 1 mm. What should be the weight per unit length of this wire so that CD remains suspended at its position? Indicate the direction of current in CD and the nature of force between the two wires.

Answer: The magnetic force is balanced by the weight of the wire: \(\dfrac Wl=\dfrac Fl=\dfrac{\mu_0I_1I_2}{2\pi d}=\dfrac{2\times10^{-7}\times12\times5}{10^{-3}}=1.2\times10^{-2}\) N/m. The current in CD is opposite to that in AB and the force between the wires is repulsive.

A small magnet of magnetic moment \(M\) is placed at a distance ‘r’ from the origin O with its axis parallel to the x-axis. A small coil of one turn and radius \(a\) is placed on the x-axis, at the same distance from the origin, with the axis of the coil coinciding with the x-axis. For what value of current in the coil does a small magnetic needle kept at the origin remain undeflected? What is the direction of the current in the coil?

Answer: This happens when the field of the bar magnet at the origin (equatorial point) equals and opposes the field of the coil at the origin (axial point): \(\dfrac{\mu_0M}{4\pi r^3}=\dfrac{\mu_0Ia^2}{2r^3}\) (for \(r\gg a\)), so \(I=\dfrac{2M}{4\pi a^2}=\dfrac{M}{2\pi a^2}\). The current must be in the sense that makes the coil’s field point opposite to the magnet’s field at O; for the arrangement in the figure (N pole of the magnet on the left) it is anticlockwise as seen from the origin.

Consider a long straight cylindrical wire of circular cross-section of radius \(a\). The current \(I\) is uniformly distributed across this cross-section. Calculate the magnetic field \(B\) in the regions \(r\lt a\) and \(r\gt a\). Plot a graph of \(B\) versus \(r\) from the centre of the wire. Calculate the ratio of the magnetic field at a point \(a/2\) above the surface of the wire to that at a point \(a/2\) below its surface.

Answer: Inside (\(r\lt a\)): the current enclosed is \(\dfrac{Ir^2}{a^2}\); Ampere’s law \(B(2\pi r)=\mu_0\dfrac{Ir^2}{a^2}\) gives \(B=\dfrac{\mu_0Ir}{2\pi a^2}\), so \(B\propto r\). At the surface (\(r=a\)), \(B=\dfrac{\mu_0I}{2\pi a}\); at the axis \(B=0\). Outside (\(r\gt a\)): \(B(2\pi r)=\mu_0I\), so \(B=\dfrac{\mu_0I}{2\pi r}\), \(B\propto\dfrac1r\).

At \(a/2\) above the surface (\(r=3a/2\)): \(B_1=\dfrac{\mu_0I}{2\pi(3a/2)}=\dfrac{\mu_0I}{3\pi a}\). At \(a/2\) below the surface (\(r=a/2\)): \(B_2=\dfrac{\mu_0I(a/2)}{2\pi a^2}=\dfrac{\mu_0I}{4\pi a}\).

\[\dfrac{B_1}{B_2}=\dfrac43\]

Name another device/machine which uses crossed electric and magnetic fields. What does this machine do and what are the functions of the magnetic and electric fields in this machine? Where do these fields exist in this machine? Write about their natures.

Answer: Name of the device: the cyclotron. It accelerates charged particles/ions. The electric field accelerates the charged particles; the magnetic field makes the particles move in a circle. The electric field exists between the dees; the magnetic field exists both inside and outside the dees. The magnetic field is uniform (constant); the electric field is oscillating (alternating) in nature.

Two identical coils P and Q, each of radius R, are lying in perpendicular planes such that they have a common centre. Find the magnitude and direction of the magnetic field at the common centre of the two coils, if they carry currents equal to \(I\) and \(\sqrt3I\) respectively.

Answer: Field due to coil P: \(B_P=\dfrac{\mu_0I}{2R}\), along the y-axis. Field due to coil Q: \(B_Q=\dfrac{\mu_0\sqrt3I}{2R}\), along the x-axis. Resultant: \(B=\sqrt{B_P^2+B_Q^2}=\dfrac{\mu_0I}{2R}\sqrt{1+3}=\dfrac{\mu_0I}{R}\). \(\tan\theta=\dfrac{B_P}{B_Q}=\dfrac1{\sqrt3}\Rightarrow\theta=30^\circ\) with the direction of \(B_Q\) (the x-axis).

(a) A point charge \(q\) moving with speed \(v\) enters a uniform magnetic field \(\vec B\) that is acting into the plane of the paper. What is the path followed by the charge \(q\) and in which plane does it move? (b) How does the path followed by the charge get affected if its velocity has a component parallel to \(\vec B\)? (c) If an electric field \(\vec E\) is also applied such that the particle continues moving along the original straight-line path, what should be the magnitude and direction of the electric field?

Answer: (a) The charge \(q\) describes a circular path, anticlockwise, in the XY plane (the plane perpendicular to \(B\)). (b) The path becomes helical. (c) The direction of the Lorentz magnetic force is along \(-Y\), so the applied electric field should be along \(+Y\). For undeflected motion the net force is zero: \(q\vec E=-q(\vec v\times\vec B)\), giving \(E=vB\).

A particle of mass \(m\) and charge \(q\) moves at right angles to a uniform magnetic field. Plot a graph showing the variation of the radius of the circular path described by it with the increase in its (a) charge (b) kinetic energy, where in each case other factors remain the same. Justify your answer.

Answer: (a) \(r=\dfrac{mv}{qB}\), so \(r\propto\dfrac1q\): the graph is a hyperbola falling with \(q\). (b) \(K=\dfrac{q^2B^2r^2}{2m}\), so \(r=\dfrac{\sqrt{2mK}}{qB}\propto\sqrt K\): the graph is a parabola rising with K.E.

A square loop of side 20 cm carrying a current of 1 A is kept near an infinite long straight wire carrying a current of 2 A in the same plane, the nearer side of the loop being 10 cm from the wire. Calculate the magnitude and direction of the net force exerted on the loop due to the current-carrying conductor.

Answer: The forces on the sides perpendicular to the wire cancel. The force on the nearer side (10 cm away) and on the farther side (30 cm away) are opposite: \(F=\dfrac{\mu_0I_1I_2l}{2\pi}\left[\dfrac1{10\times10^{-2}}-\dfrac1{30\times10^{-2}}\right]=\dfrac{4\pi\times10^{-7}\times1\times2}{2\pi}\times20\times10^{-2}\times6.67=5.33\times10^{-7}\) N. This net force is directed towards the infinitely long straight wire.

(a) Why is the magnetic field radial in a moving coil galvanometer? Explain how it is achieved. (b) A galvanometer of resistance \(G\) can be converted into a voltmeter of range (0–V) volts by connecting a resistance \(R\) in series with it. How much resistance will be required to change its range from 0 to \(V/2\)?

Answer: (a) The torque is \(NIBA\sin\theta\), where \(\theta\) is the angle between the field and the normal to the coil. For the deflection to be directly proportional to the current, \(\sin\theta\) must stay constant (\(\theta=90^\circ\)), which is possible only when the field is radial; the plane of the coil is then always parallel to the field. A radial field is obtained by making the pole pieces of the magnet concave in shape and placing a soft iron cylinder (core) between them.

(b) \(I_g=\dfrac V{R+G}\). For range \(V/2\) with series resistance \(R'\): \(I_g=\dfrac{V/2}{R'+G}\). Equating, \(2(R'+G)=R+G\), so

\[R'=\dfrac{R-G}{2}\]

Two very small identical circular loops (1) and (2), carrying equal currents \(I\), are placed vertically (with respect to the plane of the paper) with their geometrical axes perpendicular to each other, at the same distance \(x\) from the point O. Find the magnitude and direction of the net magnetic field produced at the point O.

Answer: The field of each loop at O (on its axis) is \(B_1=B_2=\dfrac{\mu_0Ir^2}{2(r^2+x^2)^{3/2}}\), and the two fields are perpendicular. Net field: \(B_0=\sqrt{B_1^2+B_2^2}=\sqrt2B=\dfrac{\sqrt2\mu_0Ir^2}{2(r^2+x^2)^{3/2}}\). For small loops (\(r\ll x\)): \(B_0=\dfrac{\sqrt2\mu_0Ir^2}{2x^3}\). Direction: \(\tan\theta=\dfrac{B_2}{B_1}=1\Rightarrow\theta=45^\circ\) with the axis of either loop.

An electron beam is deflected in a given field. How can we detect whether the given field is a uniform magnetic field or a uniform electric field?

Answer: Observe the trajectory and measure the kinetic energy of the electron beam. If the trajectory of the beam is a parabola and its K.E. changes, the beam is in an electric field. If the trajectory is circular and its K.E. remains the same, the electron is in a magnetic field.

In what way is Gauss’s law in magnetism different from that used in electrostatics? Explain briefly. The earth’s magnetic field at the equator is approximately 0.4 G. Estimate the earth’s magnetic dipole moment. Given: radius of the earth = 6400 km.

Answer: In magnetism, Gauss’s law states \(\oint\vec B\cdot d\vec S=0\); in electrostatics, \(\oint\vec E\cdot d\vec S=\dfrac q{\varepsilon_0}\). The reason is that magnetic monopoles do not exist.

Field on the equator: \(B=\dfrac{\mu_0M}{4\pi R^3}=10^{-7}\dfrac M{R^3}\), so

\[M=\dfrac{BR^3}{10^{-7}}=\dfrac{0.4\times10^{-4}\times(6400\times10^3)^3}{10^{-7}}\approx1.05\times10^{23}\ \text{A m}^2\]

An air-core solenoid having \(N\) turns and length \(l\) is carrying a current of \(I\) ampere. (i) Sketch the pattern of magnetic lines of force in the solenoid. (ii) If an iron core is inserted in it, how is its field modified? (iii) Write down the expression for the magnetic field at the centre of the solenoid along its axis.

Answer: (ii) The magnetic field lines come closer to each other (the field becomes much stronger).

(iii) \(B=\mu_0nI=\mu_0\dfrac NlI\).

A uniform magnetic field is set up along the positive x-axis. A particle of charge \(q\) and mass \(m\) moving with a velocity \(v\) enters the field at the origin in the X–Y plane such that it has velocity components both along and perpendicular to the magnetic field \(B\). Trace, giving reason, the trajectory followed by the particle. Find out the expression for the distance moved by the particle along the magnetic field in one rotation.

Answer: The particle describes a circle in the y–z plane because of the component \(v_y\) of its velocity perpendicular to the field. It also moves along the x-axis (parallel to the field) because of the component \(v_x\), which is unaffected by the field. Hence its trajectory is helical.

The distance moved along the field in one rotation is the pitch \(p=v_xT\); with the period \(T=\dfrac{2\pi m}{qB}\): \(p=\dfrac{2\pi mv_x}{qB}\).

How does a paramagnetic material behave in the presence of an external magnetic field? What happens when the temperature of a paramagnetic sample is lowered?

Answer: The individual atoms or molecules of a paramagnetic material possess a permanent magnetic dipole moment. Due to random thermal motion of the atoms, no net magnetisation is seen. In the presence of an external field \(B_0\), which is strong enough and at low temperature, the individual atomic dipole moments align and point in the direction of \(B_0\). The field lines get concentrated inside the material and the field inside is slightly increased. When placed in a non-uniform magnetic field, the bar tends to move from the weak field to the strong. As the temperature is lowered, the magnetisation increases until it reaches the saturation value, at which all the dipoles are perfectly aligned with the field.

Distinguish the magnetic properties of dia-, para- and ferromagnetic substances in terms of (i) susceptibility (ii) magnetic permeability and (iii) coercivity. Give an example of each of these materials.

Answer: The table below compares the three.

How will a dia-, a para- and a ferromagnetic material behave when kept in a non-uniform external magnetic field? Name two main characteristics of a ferromagnetic material which help to decide its suitability for making (i) a permanent magnet and (ii) an electromagnet. Which of these two characteristics should have a high or low value for each of these two types of magnets?

Answer: When kept in a non-uniform external field, a diamagnetic material moves slowly from stronger to weaker parts of the field, a paramagnetic material moves slowly from weaker to stronger parts, and a ferromagnetic material moves quickly from weaker to stronger parts.

Two main characteristics of a ferromagnetic material: retentivity and coercivity. (i) For a permanent magnet, the material should have high retentivity and high coercivity. (ii) For an electromagnet, the material should have low retentivity and low coercivity.

Two similar bars, made from two different materials P and Q, are placed one by one in a non-uniform magnetic field. It is observed that (a) the bar P tends to move from the weak to the strong field region, (b) the bar Q tends to move from the strong to the weak field region. What is the nature of the magnetic materials used for making these two bars?

Answer: Bar P is paramagnetic (it could also be ferromagnetic, if the attraction is strong); bar Q is diamagnetic.

A charged particle of mass ‘m’ and charge ‘q’ moving at a uniform velocity ‘v’ enters a uniform magnetic field ‘B’ normal to the field direction. Deduce an expression for the kinetic energy of the particle. Why does the kinetic energy of the charged particle not change when moving through the magnetic field?

Answer: The magnetic force \(qvB\) provides the centripetal force: \(qvB=\dfrac{mv^2}{r}\), so \(r=\dfrac{mv}{qB}\) and \(v=\dfrac{qBr}{m}\). Kinetic energy \(K=\tfrac12mv^2=\dfrac{q^2B^2r^2}{2m}\).

The magnetic force is always perpendicular to the velocity, so it does no work on the particle. Hence the speed, and therefore the kinetic energy, does not change.

An electron is revolving around the nucleus of an atom in an orbit of radius 0.53 Å. Calculate the equivalent magnetic moment, if the frequency of revolution of the electron is \(6.8\times10^9\) MHz.

Answer: \(f=6.8\times10^9\) MHz \(=6.8\times10^{15}\) Hz. \(M=IA=ef\,\pi r^2=1.6\times10^{-19}\times6.8\times10^{15}\times\pi(0.53\times10^{-10})^2=9.6\times10^{-24}\) A m².

A straight wire of length \(l\) is bent into a circular shape. If the wire were to carry a current of 5 A, calculate the magnetic field due to it, before bending, at a point 0.01 times the radius of the circle formed from it. Also calculate the magnetic field at the centre of the circular loop formed, for the same value of current.

Answer: (The value of the length is missing in the source; the given answers correspond to a circle of radius \(r=25\) cm, i.e. a wire of length \(2\pi r\approx1.57\) m.) Before bending, the point is at \(d=0.01r=2.5\) mm from a long straight wire: \(B_1=\dfrac{\mu_0I}{2\pi d}=\dfrac{2\times10^{-7}\times5}{2.5\times10^{-3}}=4\times10^{-4}\) T. At the centre of the loop: \(B_2=\dfrac{\mu_0I}{2r}=\dfrac{4\pi\times10^{-7}\times5}{2\times0.25}=1.256\times10^{-5}\) T.

Two insulated wires perpendicular to each other in the same plane carry equal currents as shown in the figure (one carries current upwards, the other to the left). Is there a region where the magnetic field is zero? If so, where is the region? If not, explain why the field is not zero.

Answer: Yes. In the plane of the wires, the field of each wire is perpendicular to the plane (into or out of the page). Taking the wires as axes, in two opposite quadrants the two fields point the same way and add, and in the other two quadrants (the second and the fourth) they point in opposite directions. In these two quadrants the field is zero at points equidistant from the two wires, i.e. along the line making 45° with each wire (the line \(y=-x\) for the directions drawn).

What is the net magnetic field at point O for the current distribution shown? (A circular loop of radius \(r\) with centre O carries current \(i\), and a long straight wire carrying current \(4i\) is at a distance \(2r\) from O.)

Answer: Field of the circular loop at its centre: \(B_1=\dfrac{\mu_0i}{2r}\). Field of the straight wire at distance \(2r\): \(B_2=\dfrac{\mu_0(4i)}{2\pi(2r)}=\dfrac{\mu_0i}{\pi r}\). As drawn (clockwise current in the loop, upward current in the wire) the two fields are opposite, so

\[B=B_1-B_2=\dfrac{\mu_0i}{r}\left(\dfrac12-\dfrac1\pi\right)=\dfrac{\mu_0i(\pi-2)}{2\pi r}\]

directed into the page (the direction of the loop’s field).

A galvanometer of resistance 3663 Ω gives full-scale deflection for a certain current \(I_g\). Calculate the value of the resistance of the shunt which, when joined to the galvanometer coil, will result in 1/34 of the total current passing through the galvanometer. Also find the total resistance of the galvanometer and shunt.

Answer: \(S=\dfrac{I_gG}{I-I_g}\) with \(I_g=\dfrac I{34}\): \(S=\dfrac{3663}{33}=111\ \Omega\). Total resistance \(=\dfrac{GS}{G+S}=\dfrac{3663\times111}{3774}\approx107.7\ \Omega\).

1. Magnetic field:

dB =μ₀ IdlSinθ/4πr²

μ₀ =4π x 10^-7 Tm/A 

*Direction of dB can be found by using Maxwell’s Right hand thumb rule.+

3. Applications :

5. Applications

B= μ₀ N I / 2πr

N= Total no. of turns.