CBSE Class 12 Physics Magnetic Effects of Electric Current Objective Questions

CBSE Class 12 Physics Magnetic Effects of Electric Current Objective Questions. 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.

OBJECTIVES
CLASS 12 PHYSICS : MAGNETIC EFFECTS OF ELECTRIC CURRENT

1. Two circular coils P and Q are made from similar wires but radius of Q is twice that of P. What should be the value of potential difference across them so that the magnetic induction at their centre may be the same?
(a) V_q = 2 V_p
(b) V_q = 3V_p
(c) V_q = 4V_p
(d) V_q = 1/4 V_p

Answer: (c) V_q = 4V_p

1. Each coil is one turn of the same kind of wire, so its resistance is proportional to its length, that is, to its radius: R_q = 2R_p.
2. Current I = V/R and magnetic induction at the centre B = μ₀I/(2r). So B is proportional to V/(R r), that is, to V/r².
3. For equal B: V_q/r_q² = V_p/r_p², so V_q = V_p × (2)² = 4V_p.
4. Hence, the correct option is (c).

Teacher's Note:
a) Both the current and the field formula depend on the radius, so the effect appears twice.
b) A longer wire has more resistance, which reduces the current.

2. The radius of the circular path or helical path followed by the test change q₀ moving in magnetic field B with velocity v is
(a) \( \frac{m v sin θ}{q0 B} \)
(b) \( \frac{m v cos θ}{q0 B} \)
(c) \( \frac{m v}{q0 B} \)
(d) \( \frac{m v tan θ}{q0 B} \)

Answer: (a) \( \frac{m v sin θ}{q0 B} \)

1. Only the component of velocity perpendicular to B, v sin θ, produces the circular motion; θ is the angle between v and B.
2. The magnetic force provides the centripetal force: q₀(v sin θ)B = m(v sin θ)²/r.
3. r = m v sin θ/(q₀B).
4. Hence, the correct option is (a).

Teacher's Note:
a) The component along B is unaffected and gives the pitch of the helix.
b) For θ = 90° the path is a circle of radius mv/(qB).

3. An electron accelerated through a potential difference V enters into a uniform transverse magnetic field and experiences a force F. If the accelerating potential is increased to 2 V, the electron in the same magnetic field will experience a force :
(a) F
(b) F/2
(c) √2 F
(d) 2F

Answer: (c) √2 F

1. The speed gained: (1/2)mv² = eV, so v is proportional to √V.
2. The magnetic force F = evB is proportional to v, so F is proportional to √V.
3. For potential 2V: F' = √2 F.
4. Hence, the correct option is (c).

Teacher's Note:
a) The force depends on the speed, not on the kinetic energy.
b) Doubling the potential increases the speed by √2 only.

4. A charged particle moving in a uniform magnetic field penetrates a layer of lead and loses one half of its kinetic energy. The ratio of radius of curvature to the original radius is
(a) 2
(b) √2
(c) \( \frac{1}{2} \)
(d) \( \frac{1}{\sqrt{2}} \)

Answer: (d) \( \frac{1}{\sqrt{2}} \)

1. The radius r = mv/(qB) is proportional to the speed v.
2. Kinetic energy is halved, so v² is halved and v becomes v/√2.
3. r'/r = 1/√2.
4. Hence, the correct option is (d).

Teacher's Note:
a) Kinetic energy is proportional to v², but the radius is proportional to v.
b) The path becomes more tightly curved after passing through the lead.

5. Two straight long conductors AOB and COD are perpendicular to each other and carry currents i₁ and i₂. The magnitude of the magnetic induction at a point P at a distance d from the point O in a direction perpendicular to the plane ABCD is
(a) \( \frac{μ0}{2πd} \) (i₁ + i₂)
(b) \( \frac{μ0}{4πd} \) (i₁ - i₂)
(c) \( \frac{μ0}{2πd} \) (i₁² + i₂²)^1/2
(d) \( \frac{μ0}{2πd} \) \( \left( \frac{i_1 i_2}{i_1 + i_2} \right) \)

Answer: (c) \( \frac{μ0}{2πd} \) (i₁² + i₂²)^1/2

1. The point P is at the distance d from each conductor, along the line perpendicular to both.
2. Field due to each conductor: B₁ = μ₀i₁/(2πd) and B₂ = μ₀i₂/(2πd).
3. The conductors are perpendicular, so B₁ and B₂ are perpendicular to each other.
4. B = √(B₁² + B₂²) = (μ₀/2πd)(i₁² + i₂²)^1/2.
5. Hence, the correct option is (c).

Teacher's Note:
a) Add perpendicular fields by the Pythagoras rule, not by simple addition.
b) Each long wire gives the field μ₀i/(2πd).

6. A current of 10 amp. is flowing in a wire of length 1.5 metre. A force of 15 newton acts on it when it is placed in a uniform magnetic field of 2 tesla. The angle between the magnetic field and the direction of current is
(a) 30º
(b) 45º
(c) 60º
(d) 90º

Answer: (a) 30º

1. The force on a current-carrying wire is F = BIL sin θ.
2. 15 = 2 × 10 × 1.5 × sin θ = 30 sin θ.
3. sin θ = 0.5, so θ = 30°.
4. Hence, the correct option is (a).

Teacher's Note:
a) Maximum force is BIL = 30 N, at 90°.
b) The force of 15 N is half the maximum, so sin θ = 1/2.

7. A particle carrying a charge equal to 100 times the charge on an electron is rotating per second in a circular path of radius 0.8 m. The value of the magnetic field produced at the centre will be (μ₀ = permeability constant).
(a) 10^-7/μ₀
(b) 10^-17 μ₀
(c) 10^-6 / μ₀
(d) 10^-7 μ₀

Answer: (b) 10^-17 μ₀

1. The charge q = 100e = 100 × 1.6 × 10⁻¹⁹ C = 1.6 × 10⁻¹⁷ C, and it makes 1 revolution per second.
2. Equivalent current I = qf = 1.6 × 10⁻¹⁷ × 1 = 1.6 × 10⁻¹⁷ A.
3. B = μ₀I/(2r) = μ₀ × 1.6 × 10⁻¹⁷/(2 × 0.8) = 10⁻¹⁷ μ₀.
4. Hence, the correct option is (b).

Teacher's Note:
a) A revolving charge acts as a circular current I = qf.
b) The answer is in terms of μ₀, so do not substitute its numerical value.

8. The magnetic field at the centre O of the circular portion of the current carrying wire of radius 3 cm is
(a) (8π/3) × 10^-5 T
(b) (8π/3) × 10^-4 T
(c) 2π × 10^-5 T
(d) 2π × 10^-4 T

Answer: (c) 2π × 10^-5 T

1. The straight parts AB and DE lie along lines passing through O, so they produce no field at O.
2. The circular portion subtends an angle 2π - π/2 = 3π/2 at O.
3. B = (μ₀/4π) × I × θ/r = 10⁻⁷ × 4 × (3π/2)/(3 × 10⁻²) = 2π × 10⁻⁵ T.
4. Hence, the correct option is (c).

Teacher's Note:
a) Field at the centre of an arc is (μ₀/4π)(Iθ/r).
b) A straight wire pointing directly at the point gives zero field, since dl × r is zero.

9. A long wire carries a current of 20 A along the axis of a solenoid. The field due to the solenoid is 4 mT. the resultant field at a point 3 mm from the solenoid axis is
(a) 1.33 m T
(b) 4.2 mT
(c) 2.1 m T
(d) 8.4 m T

Answer: (b) 4.2 mT

1. Field of the long wire at 3 mm: B_w = μ₀I/(2πr) = (2 × 10⁻⁷ × 20)/(3 × 10⁻³) = 1.33 × 10⁻³ T = 1.33 mT.
2. The wire's field is circular around the axis, so it is perpendicular to the solenoid field along the axis.
3. B = √(4² + 1.33²) = √17.8 = 4.2 mT.
4. Hence, the correct option is (b).

Teacher's Note:
a) The two fields are at right angles, so they add as vectors.
b) The field of the solenoid is uniform and axial.

10. A circular current carrying coil has a radius R. The distance from the centre of the coil on the axis where the magnetic induction will be (1/8)th of its value at the centre of the coil is
(a) R √3
(b) 2R √3
(c) R/√3
(d) 2R/√3

Answer: (a) R √3

1. On the axis, B = μ₀IR²/[2(R² + x²)^3/2], and at the centre B₀ = μ₀I/(2R).
2. B/B₀ = R³/(R² + x²)^3/2 = 1/8.
3. (R² + x²)^3/2 = 8R³, so R² + x² = 4R².
4. x² = 3R², so x = R√3.
5. Hence, the correct option is (a).

Teacher's Note:
a) Take the cube root of both sides carefully: the cube root of 8R³ is 2R, so (R² + x²)^(1/2) = 2R.
b) Check: with x = R√3 the distance from the rim is 2R.

11. A proton moving with a constant velocity passes through a region of space without any change in its velocity. If E and B represent the electric and magnetic fields respectively, identify wrong statement
(a) E = 0, B = 0
(b) E = 0, B ≠ 0
(c) E ≠ 0, B = 0
(d) E ≠ 0, B ≠ 0

Answer: (c) E ≠ 0, B = 0

1. If there is no change in velocity, the net force on the proton is zero.
2. With E = 0 and B = 0 there is no force; with E = 0 and B ≠ 0 the force is zero if the velocity is parallel to B.
3. With E ≠ 0 and B ≠ 0 the net force is zero if qE = qvB with the forces opposite.
4. With E ≠ 0 and B = 0 there is a force qE, which changes the velocity. So this statement is wrong.
5. Hence, the correct option is (c).

Teacher's Note:
a) The question asks for the wrong statement, so look for the case where a force must exist.
b) An electric field always exerts a force on a charge, whatever its velocity.

12. An electron is moving in a circle of radius r in a uniform magnetic field B. Suddenly the field is reduced to B/2. The radius of the circle now becomes
(a) r/2
(b) r/4
(c) 2 r
(d) 4 r

Answer: (c) 2 r

1. The radius is r = mv/(eB). The speed of the electron does not change when the field changes, since the magnetic force does no work.
2. r is inversely proportional to B, so for B/2 the radius is 2r.
3. Hence, the correct option is (c).

Teacher's Note:
a) A magnetic field can change the direction of the velocity but not its magnitude.
b) Weaker field means a larger circle.

13. A proton and an α-particle, moving with the same kinetic energy, enters a uniform magnetic field normally. The radii of their circular paths will be in the ratio
(a) 1 : 1
(b) 2 : 1
(c) 1 : 2
(d) 4 : 1

Answer: (a) 1 : 1

1. For a given kinetic energy K, the momentum p = √(2mK), so r = p/(qB) = √(2mK)/(qB).
2. r is proportional to √m/q. For the proton: √1/1 = 1. For the α-particle: √4/2 = 1.
3. r_p : r_α = 1 : 1.
4. Hence, the correct option is (a).

Teacher's Note:
a) An α-particle has four times the mass and twice the charge of a proton.
b) Compare √m/q for different particles when the kinetic energy is equal.

14. A beam of electrons is accelerated through a potential difference V. It is then passed normally through a uniform magnetic field where it moves in a circle of radius r. It would have moved in a circle of radius 2r if it were initially accelerated through a potential difference
(a) √2 V
(b) 2V
(c) 2√2 V
(d) 4V

Answer: (d) 4V

1. eV = (1/2)mv², so v = √(2eV/m).
2. r = mv/(eB) = (1/B)√(2mV/e), so r is proportional to √V.
3. For radius 2r, √V' = 2√V, so V' = 4V.
4. Hence, the correct option is (d).

Teacher's Note:
a) Radius goes as the square root of the accelerating potential.
b) To double the radius, quadruple the potential.

15. A proton and an α-particle, accelerated through the same potential difference, enter a region of uniform magnetic field normally. If the radius of the proton orbit is 10 cm then the radius of of the α-orbit is
(a) 10 cm
(b) 10√2 cm
(c) 20 cm
(d) 5√2 cm

Answer: (b) 10√2 cm

1. After acceleration through V, v = √(2qV/m), so r = mv/(qB) = (1/B)√(2mV/q).
2. r is proportional to √(m/q). For the proton the value is 1, and for the α-particle √(4/2) = √2.
3. r_α = 10√2 cm.
4. Hence, the correct option is (b).

Teacher's Note:
a) At the same potential difference the kinetic energy is qV, which is different for the two particles.
b) Do not confuse this with the case of equal kinetic energy.

16. Two particles X and Y having equal charge, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii R₁ and R₂ respectively. The ratio of the mass of X to that of Y is
(a) (R₁ / R₂)^1/2
(b) R₂ / R₁
(c) (R₁ / R₂)²
(d) R₁ / R₂

Answer: (c) (R₁ / R₂)²

1. For the same charge and potential difference, r = (1/B)√(2mV/q), so r² is proportional to m.
2. m_X/m_Y = R₁²/R₂² = (R₁/R₂)².
3. Hence, the correct option is (c).

Teacher's Note:
a) The radius is proportional to the square root of the mass.
b) Square both sides to remove the root.

17. A particle of charge +q and mass m moving under the influence of a uniform electric field Ei and a uniform magnetic field B k follows a trajectory from P to Q as shown in the figure. The velocities at P and Q are vi and -2vj. Which of the following statements is incorrect
(a) E = \( \frac{3}{4} \) \( \left( \frac{m v^2}{qa} \right) \)
(b) Rate of work done by the electric field at P is \( \frac{3}{4} \) \( \left( \frac{m v^3}{a} \right) \)
(c) Rate of work done by the electric field at P is zero
(d) Rate of work done by both the fields at Q is zero

Answer: (c) Rate of work done by the electric field at P is zero

1. The magnetic force does no work, so the work done by the electric field from P to Q equals the change in kinetic energy: qE(2a) = (1/2)m(2v)² - (1/2)mv² = (3/2)mv².
2. So E = 3mv²/(4qa). Statement (a) is correct.
3. At P the rate of work by the electric field = qE × v = q × (3mv²/4qa) × v = 3mv³/(4a). Statement (b) is correct and statement (c) is incorrect.
4. At Q the velocity is along -y, perpendicular to the electric force (along x), and the magnetic force never does work, so the total rate is zero. Statement (d) is correct.
5. Hence, the incorrect statement is option (c).

Teacher's Note:
a) The question asks for the incorrect statement.
b) The magnetic force is always perpendicular to the velocity, so its power is zero.
c) Power of a force is F · v.

18. A horizontal wire of length 10 cm and mass 0.3 g carries a current of 5 A. The magnitude of the magnetic field which can support the weight of the wire is (g = 10 m/s²)
(a) 3 × 10^-3 T
(b) 6 × 10^-3 T
(c) 3 × 10^-4 T
(d) 6 × 10^-4 T

Answer: (b) 6 × 10^-3 T

1. For the wire to be supported, the magnetic force must balance the weight: BIL = mg.
2. B = mg/(IL) = (0.3 × 10⁻³ × 10)/(5 × 0.1) = 3 × 10⁻³/0.5.
3. B = 6 × 10⁻³ T.
4. Hence, the correct option is (b).

Teacher's Note:
a) Convert grams to kilograms and centimetres to metres before substituting.
b) The field must be perpendicular to the wire and directed so that the force is upward.

19. A 2 MeV proton (mass = 1.6 × 10^-27 kg) is moving perpendicular to a uniform magnetic field of 2.5 T. The magnetic force on the proton is
(a) 2.5 × 10^-10 N
(b) 8 × 10^-11 N
(c) 2.5 × 10^-11 N
(d) 8 × 10^-12 N

Answer: (d) 8 × 10^-12 N

1. Kinetic energy K = 2 MeV = 2 × 1.6 × 10⁻¹³ J = 3.2 × 10⁻¹³ J.
2. v = √(2K/m) = √(2 × 3.2 × 10⁻¹³/1.6 × 10⁻²⁷) = √(4 × 10¹⁴) = 2 × 10⁷ m/s.
3. F = qvB = 1.6 × 10⁻¹⁹ × 2 × 10⁷ × 2.5 = 8 × 10⁻¹² N.
4. Hence, the correct option is (d).

Teacher's Note:
a) Convert MeV to joules using 1 eV = 1.6 × 10⁻¹⁹ J.
b) Perpendicular motion gives the maximum force, qvB.

20. A portion of a long straight wire, carrying a current I, is bent in the form of a semicircle of radius r as shown in the figure. The magnetic field at the centre O of the semicircle, in tesla, is
(a) \( \frac{π I}{r} \)×10^-7
(b) \( \frac{π I}{r} \)
(c) \( \frac{π I}{r} \)×10⁷
(d) \( \frac{π I}{2r} \)

Answer: (a) \( \frac{π I}{r} \)×10^-7

1. The straight portions lie along a line through the centre O, so they give no field at O.
2. Field due to a semicircle at its centre: B = μ₀I/(4r) = (μ₀/4π)(πI/r).
3. With μ₀/4π = 10⁻⁷ T m/A, B = (πI/r) × 10⁻⁷ T.
4. Hence, the correct option is (a).

Teacher's Note:
a) A semicircle gives half of the field of a full circle, μ₀I/(2r).
b) Express the answer using μ₀/4π = 10⁻⁷ to match the option.

21. A long straight conductor, carrying a current I, is bent into the shape shown in the figure. The radius of the circular loop is r. The magnetic field at the centre of the loop is
(a) \( \frac{μ0 I}{2r} \) \( \left( 1 + \frac{1}{\pi} \right) \) into the page
(b) \( \frac{μ0 I}{2r} \) \( \left( 1 - \frac{1}{\pi} \right) \) out of the page
(c) \( \frac{μ0 I}{2r} \) \( \left( 1 - \frac{1}{\pi} \right) \) into the page
(d) \( \frac{μ0 I}{2r} \) \( \left( 1 + \frac{1}{\pi} \right) \) out of the page

Answer: (c) \( \frac{μ0 I}{2r} \) \( \left( 1 - \frac{1}{\pi} \right) \) into the page

1. The current goes once round the loop, in the clockwise sense, so the loop gives B₁ = μ₀I/(2r) into the page.
2. The straight conductor passes at a distance r below the centre, and the current to the right gives B₂ = μ₀I/(2πr) out of the page at the centre.
3. The two fields are opposite, so B = B₁ - B₂ = (μ₀I/2r)(1 - 1/π) into the page.
4. Hence, the correct option is (c).

Teacher's Note:
a) The straight part is a very long wire touching the loop, so its field is μ₀I/(2πr).
b) Use the right-hand rule to decide the directions; the loop dominates because 1/π is less than 1.

22. A rectangular loop carrying a current i is situated near a long straight wire such that the wire is parallel to one of the sides of the loop and is in the plane of the loop. If a steady current I is established in the wire as shown in the figure, the loop will
(a) rotate about an axis parallel to the wire
(b) move away from the wire
(c) move towards the wire
(d) remain stationary

Answer: (c) move towards the wire

1. The side of the loop nearest the wire carries a current in the same direction as the current I in the wire (both upward), so they attract.
2. The side of the loop farthest from the wire carries a current in the opposite direction, so it is repelled, but the force is weaker because it is farther.
3. The forces on the top and bottom sides cancel. The net force is towards the wire.
4. Hence, the loop moves towards the wire and the correct option is (c).

Teacher's Note:
a) Parallel currents attract and antiparallel currents repel.
b) The nearer side has the stronger field, so the attraction wins.

23. A square coil of side a carries a current I. The magnetic field at the centre of the coil is
(a) \( \frac{μ0 I}{aπ} \)
(b) \( \frac{√2 μ0 I}{aπ} \)
(c) \( \frac{μ0 I}{\sqrt{2}\,a\pi} \)
(d) \( \frac{2√2 μ0 I}{aπ} \)

Answer: (d) \( \frac{2√2 μ0 I}{aπ} \)

1. The centre is at a distance d = a/2 from each side, and each side subtends 45° on either side of the perpendicular from the centre.
2. Field due to one side: B₁ = (μ₀I/4πd)(sin 45° + sin 45°) = (μ₀I/(4π(a/2))) × √2 = √2 μ₀I/(2πa).
3. All four sides give fields in the same direction, so B = 4B₁ = 2√2 μ₀I/(πa).
4. Hence, the correct option is (d).

Teacher's Note:
a) Use the finite wire formula (μ₀I/4πd)(sin θ₁ + sin θ₂).
b) Multiply by four, since all sides contribute in the same direction.

24. A circular current -carrying coil has a radius R. The distance from the centre of the coil, on the axis, where B will be 1/8th of its value at the centre of the coil is
(a) R / √3
(b) √3 R
(c) 2√3 R
(d) \( \frac{2R}{√3} \)

Answer: (b) √3 R

1. On the axis, B = μ₀IR²/[2(R² + x²)^3/2] and B₀ = μ₀I/(2R).
2. B/B₀ = R³/(R² + x²)^3/2 = 1/8, so (R² + x²)^3/2 = 8R³.
3. R² + x² = 4R², so x = √3 R.
4. Hence, the correct option is (b).

Teacher's Note:
a) The cube root of 8 is 2.
b) This is the same relation as in Question 10.

25. In a figure AB is a long straight wire carrying a current of 20 A and CDFG is a rectangular loop of size 20 cm × 9 cm carrying a current of 10 A. The edge CG is parallel to AB, at a distance of 1 cm from it. The force exerted on the loop by the magnetic field of the wire is
(a) 3.6 × 10^-4 N towards left
(b) 3.6 × 10^-4 N towards right
(c) 7.2 × 10^-4 N towards right
(d) 7.2 × 10^-4 N towards left

Answer: (d) 7.2 × 10^-4 N towards left

1. The sides CD and GF are at equal distances and carry opposite currents, so their forces cancel.
2. Side CG (1 cm from the wire) carries a current parallel to the 20 A current, so it is attracted: F₁ = μ₀I₁I₂L/(2πd) = (2 × 10⁻⁷ × 20 × 10 × 0.2)/0.01 = 8 × 10⁻⁴ N towards the wire (left).
3. Side DF is at 10 cm from the wire and carries an antiparallel current, so it is repelled: F₂ = (2 × 10⁻⁷ × 20 × 10 × 0.2)/0.1 = 0.8 × 10⁻⁴ N (right).
4. Net force = 8 × 10⁻⁴ - 0.8 × 10⁻⁴ = 7.2 × 10⁻⁴ N towards the left.
5. Hence, the correct option is (d).

Teacher's Note:
a) The nearer side dominates because the force is inversely proportional to the distance.
b) The current in the side nearer the wire runs in the same direction as in the wire, so the force is attractive.

26. In the figure x, y and z are long straight wires. The magnetic force on 25 cm length of y is
(a) 2 × 10^-4 N towards right
(b) 4 × 10^-4 N towards right
(c) 2 × 10^-4 N towards left
(d) 4 × 10^-4 N towards left

Answer: (b) 4 × 10^-4 N towards right

1. Currents: x = 30 A upward, y = 10 A downward, z = 20 A upward. Distances from y: x at 3 cm on its left and z at 10 cm on its right.
2. x and y carry opposite currents, so y is repelled by x, towards the right: F₁ = (2 × 10⁻⁷ × 30 × 10 × 0.25)/0.03 = 5 × 10⁻⁴ N.
3. z and y also carry opposite currents, so y is repelled by z, towards the left: F₂ = (2 × 10⁻⁷ × 20 × 10 × 0.25)/0.10 = 1 × 10⁻⁴ N.
4. Net force = 5 × 10⁻⁴ - 1 × 10⁻⁴ = 4 × 10⁻⁴ N towards the right.
5. Hence, the correct option is (b).

Teacher's Note:
a) Opposite currents repel; the direction of each force is away from the other wire.
b) Subtract the forces when they act in opposite directions.

27. A current of I ampere flows in a wire forming a circular arc of radius r metres subtending an angle θ at the centre as shown. The magnetic field at the centre O in tesla is
(a) \( \frac{μ0Iθ}{4πr} \)
(b) \( \frac{μ0Iθ}{2πr} \)
(c) \( \frac{μ0Iθ}{2r} \)
(d) \( \frac{μ0Iθ}{4r} \)

Answer: (a) \( \frac{μ0Iθ}{4πr} \)

1. By the Biot-Savart law, for an arc of radius r, each element dl contributes dB = (μ₀/4π) I dl/r².
2. The arc length is rθ, so B = (μ₀/4π) × I × rθ/r² = μ₀Iθ/(4πr).
3. Hence, the correct option is (a).

Teacher's Note:
a) For a full circle θ = 2π, which gives the known result μ₀I/(2r).
b) The angle must be in radians.

28. In the given loop, the magnetic field at the centre O is
(a) \( \frac{μ0I}{4} \) \( \left( \frac{r_1 + r_2}{r_1 r_2} \right) \) out of the page
(b) \( \frac{μ0I}{4} \) \( \left( \frac{r_1 + r_2}{r_1 r_2} \right) \) into the page
(c) \( \frac{μ0I}{4} \) \( \left( \frac{r_1 - r_2}{r_1 r_2} \right) \) out of the page
(d) \( \frac{μ0I}{2} \) \( \left( \frac{r_1 - r_2}{r_1 r_2} \right) \) out of the page

Answer: (b) \( \frac{μ0I}{4} \) \( \left( \frac{r_1 + r_2}{r_1 r_2} \right) \) into the page

1. The straight parts of the wire pass through O, so they produce no field there.
2. The smaller semicircle (radius r₂) gives B₂ = μ₀I/(4r₂) and the larger semicircle (radius r₁) gives B₁ = μ₀I/(4r₁).
3. The current goes clockwise round both semicircles, so both fields point into the page and they add.
4. B = (μ₀I/4)(1/r₁ + 1/r₂) = (μ₀I/4)(r₁ + r₂)/(r₁r₂), into the page.
5. Hence, the correct option is (b).

Teacher's Note:
a) A semicircle gives half of the field of a full circle.
b) Fields from currents circulating the same way add.

29. A proton and an α-particle enter a uniform magnetic field perpendicularly, with the same speed. If the proton takes 25 microseconds to make 5 revolutions, the periodic time for the α-particle would be
(a) 50 μs
(b) 25 μs
(c) 10 μs
(d) 5 μs

Answer: (c) 10 μs

1. Periodic time of the proton T_p = 25/5 = 5 μs.
2. T = 2πm/(qB), which is proportional to m/q and does not depend on speed.
3. m/q for the α-particle is (4/2) = 2 times that of the proton, so T_α = 2 × 5 = 10 μs.
4. Hence, the correct option is (c).

Teacher's Note:
a) The time period of circular motion in a magnetic field is independent of speed and radius.
b) The period is proportional to mass divided by charge.

30. A proton, a deuteron and an α-particle, having the same kinetic energy, are moving in circular trajectories in a constant magnetic field. If r_p, r_d and r_α denote, respectively, the radii of the trajectories of these particles, then
(a) r_α = r_p < r_d
(b) r_α > r_d > r_p
(c) r_α = r_d > r_p
(d) r_p = r_d < r_α

Answer: (a) r_α = r_p < r_d

1. For the same kinetic energy, r = √(2mK)/(qB), so r is proportional to √m/q.
2. Proton: √1/1 = 1. Deuteron (mass 2, charge 1): √2/1 = 1.41. α-particle (mass 4, charge 2): √4/2 = 1.
3. So r_α = r_p and both are less than r_d.
4. Hence, the correct option is (a).

Teacher's Note:
a) The deuteron has twice the mass of the proton and the same charge.
b) Take the ratio √m/q for each particle to compare.