CBSE Class 12 Physics Moving Charges and Magnetism MCQs Set 06

Answer: (b) 1.6 × 10⁻¹³ N

Explanation:
1. F = qvB sin θ, with θ = 90°
2. F = 1.6 × 10⁻¹⁹ × 2 × 10⁶ × 0.5 = 1.6 × 10⁻¹³ N

Teacher's Note:
1. When v is perpendicular to B, sin θ = 1 and the force is maximum.
2. The charge of a proton equals e = 1.6 × 10⁻¹⁹ C.

Answer: (d) Kinetic energy

Explanation:
1. The magnetic force is always perpendicular to the velocity, so it does no work.
2. The speed and hence the kinetic energy stay constant, while the direction of v, the momentum and the acceleration keep changing.

Teacher's Note:
1. Velocity, momentum and acceleration are vectors; their directions change on a circle.
2. Speed and kinetic energy are scalars that stay fixed.

Answer: (b) 17 cm

Explanation:
1. r = \( \frac{mv}{qB} \) = \( \frac{9.1 \times 10^{-31} \times 3 \times 10^{7}}{1.6 \times 10^{-19} \times 1 \times 10^{-3}} \)
2. r = 0.17 m = 17 cm

Teacher's Note:
1. Keep track of powers of ten: 10⁻³¹ × 10⁷ / (10⁻¹⁹ × 10⁻³) = 10⁻².
2. A stronger field or a slower particle gives a smaller circle.

Answer: (a) 1 : 2

Explanation:
1. r = \( \frac{mv}{qB} \), so for the same v and B, r ∝ \( \frac{m}{q} \).
2. Proton: m/q = m/e. Alpha particle: m/q = 4m/2e = 2(m/e).
3. r_p : r_α = 1 : 2

Teacher's Note:
1. An alpha particle has four times the mass and twice the charge of a proton.
2. Compare m/q, not m alone.

Answer: (c) a helix

Explanation:
1. The component v cos 40° along B is unaffected, so the particle drifts steadily along the field.
2. The component v sin 40° perpendicular to B produces circular motion.
3. Together they give a helix with its axis along B.

Teacher's Note:
1. A parabola is the path of a charge in a uniform electric field, not a magnetic one.
2. A circle occurs only when v is exactly perpendicular to B.

Answer: (c) 0.2 N

Explanation:
1. F = IlB sin θ = 4 × 0.5 × 0.2 × sin 30°
2. F = 0.4 × 0.5 = 0.2 N

Teacher's Note:
1. 0.4 N is the force when the wire is perpendicular to the field.
2. θ is the angle between the wire (current) and B.

Answer: (d) 6.28 × 10⁻⁴ T

Explanation:
1. B = \( \frac{\mu_0 N I}{2R} \) = \( \frac{4\pi \times 10^{-7} \times 50 \times 2}{2 \times 0.1} \)
2. B = 2π × 10⁻⁴ = 6.28 × 10⁻⁴ T

Teacher's Note:
1. Do not forget to multiply by N for a coil of many turns.
2. Use the radius, not the diameter, in the formula.

Answer: (a) 4 × 10⁻⁵ T

Explanation:
1. B = \( \frac{\mu_0 I}{2\pi r} \) = \( \frac{2 \times 10^{-7} \times 10}{0.05} \)
2. B = 4 × 10⁻⁵ T

Teacher's Note:
1. μ₀/2π = 2 × 10⁻⁷ T m A⁻¹ makes this quick to compute.
2. 1.26 × 10⁻⁴ T comes from wrongly using μ₀I/2r, the formula for the centre of a loop.

Answer: (b) 2.51 × 10⁻³ T

Explanation:
1. Turns per unit length: n = \( \frac{500}{0.5} \) = 1000 m⁻¹
2. B = μ₀nI = 4π × 10⁻⁷ × 1000 × 2 = 2.51 × 10⁻³ T

Teacher's Note:
1. n is turns per metre, so divide the total turns by the length in metres.
2. The field inside a long solenoid does not depend on its radius.

Answer: (c) 2 × 10⁻⁵ N m⁻¹, attractive

Explanation:
1. f = \( \frac{\mu_0 I_1 I_2}{2\pi d} \) = \( \frac{2 \times 10^{-7} \times 2 \times 5}{0.1} \) = 2 × 10⁻⁵ N m⁻¹
2. Currents in the same direction attract, so the answer is 2 × 10⁻⁵ N m⁻¹, attractive.

Teacher's Note:
1. Parallel currents attract and antiparallel currents repel.
2. The force on each wire has the same magnitude.

Answer: (a) 0.04 N m

Explanation:
1. With the plane parallel to B, the normal to the coil is perpendicular to B, so θ = 90°.
2. τ = NIAB sin 90° = 100 × 0.5 × 2 × 10⁻³ × 0.4 = 0.04 N m

Teacher's Note:
1. θ is measured from the normal to the coil, not from its plane.
2. The torque is zero when the plane of the coil is perpendicular to B.

Answer: (b) towards the bottom of the page

Explanation:
1. For a positive charge, \( \vec{v} \times \vec{B} \) with v to the right and B into the page points towards the top of the page.
2. The electron is negative, so the force is reversed: towards the bottom of the page.

Teacher's Note:
1. Always reverse the direction found by the right-hand rule when the charge is negative.
2. The force is perpendicular to both v and B, so it lies in the plane of the page here.

Answer: (d) 0.05 Ω

Explanation:
1. S = \( \frac{I_g G}{I - I_g} \) = \( \frac{0.005 \times 50}{5 - 0.005} \)
2. S = \( \frac{0.25}{4.995} \) ≈ 0.05 Ω

Teacher's Note:
1. The shunt is connected in parallel and must be much smaller than G.
2. Convert mA to A before using the formula.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. A force perpendicular to the velocity does no work (W = F · s = 0).
2. With no work done, the kinetic energy stays constant. Both statements are correct.
3. The Reason explains the Assertion.

Teacher's Note:
1. Only the direction of motion changes in a magnetic field.
2. An electric field, in contrast, can change a particle's kinetic energy.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Each wire lies in the field of the other, and the force F = I(l × B) on it points towards the other wire. The Assertion is correct.
2. Like charges do repel, so the Reason is also correct.
3. The attraction between currents is a magnetic effect and has nothing to do with the electrostatic force between charges, so the Reason does not explain the Assertion.

Teacher's Note:
1. The wires are electrically neutral; only the currents interact.
2. Note the contrast: like currents attract, like charges repel.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. A shunt S in parallel with G gives a combined resistance GS/(G + S), which is less than S.
2. Since S is small, the ammeter resistance is very small. Both statements are correct.
3. The parallel shunt is exactly why the resistance is low, so the Reason explains the Assertion.

Teacher's Note:
1. A low resistance lets the ammeter be put in series without changing the current much.
2. An ideal ammeter has zero resistance.

Answer: (c) A is true, but R is false.

Explanation:
1. Inside a long solenoid, B = μ₀nI, the same at all points away from the ends. The Assertion is correct.
2. Outside a long solenoid the field is nearly zero, so the Reason is false.

Teacher's Note:
1. The field lines are crowded inside and spread very thinly outside.
2. This is why a solenoid is used to produce a uniform field.

Answer: (d) A is false, but R is true.

Explanation:
1. The forces on opposite sides of the loop are equal and opposite, so the net force is zero. The Assertion is false.
2. These forces can form a couple, giving τ = NIAB sin θ. The Reason is true.

Teacher's Note:
1. Zero net force does not mean zero torque.
2. A loop in a non-uniform field can feel a net force too.

Answer: (d) A is false, but R is true.

Explanation:
1. Current sensitivity NAB/k doubles when N doubles, so the Reason is true.
2. Voltage sensitivity NAB/(kR_G) involves the coil resistance, which also roughly doubles with twice the wire. The voltage sensitivity therefore stays about the same, and the Assertion is false.

Teacher's Note:
1. Voltage sensitivity = current sensitivity ÷ coil resistance.
2. Increasing B or reducing k raises both sensitivities.

Answer: (c) 20.9 cm

Explanation:
1. r = \( \frac{m_p v}{eB} \) = \( \frac{1.67 \times 10^{-27} \times 2 \times 10^{6}}{1.6 \times 10^{-19} \times 0.1} \)
2. r = 0.209 m ≈ 20.9 cm

Teacher's Note:
1. The magnetic force qvB supplies the centripetal force mv²/r.
2. Use SI units throughout and convert to cm only at the end.

Answer: (a) 6.6 × 10⁻⁷ s

Explanation:
1. T = \( \frac{2\pi m_p}{eB} \) = \( \frac{2 \times 3.14 \times 1.67 \times 10^{-27}}{1.6 \times 10^{-19} \times 0.1} \)
2. T ≈ 6.6 × 10⁻⁷ s

Teacher's Note:
1. You can also use T = 2πr/v = 2π × 0.209/(2 × 10⁶).
2. The period does not involve the speed directly.

Answer: (b) the radius doubles and the period stays the same

Explanation:
1. r = mv/qB is proportional to v, so the radius doubles.
2. T = 2πm/qB does not contain v, so the period is unchanged.
3. Answer: the radius doubles and the period stays the same.

Teacher's Note:
1. A faster particle covers a bigger circle in the same time.
2. This speed-independence of T is an important result.

Answer: (d) zero

Explanation:
1. The magnetic force is always perpendicular to the displacement of the proton.
2. So the work done is zero at every instant, and over the whole revolution.

Teacher's Note:
1. The speed of the proton stays constant.
2. Work needs a force component along the displacement.

Answer: (c) 41.8 cm

Explanation:
1. For an alpha particle, m = 4m_p and q = 2e, so r_α = \( \frac{4m_p v}{2eB} \) = 2r_p
2. r_α = 2 × 20.9 = 41.8 cm

Teacher's Note:
1. The ratio m/q decides the radius at a given speed and field.
2. An alpha particle is a helium nucleus: two protons and two neutrons.

Answer: (b) 0.1 Ω

Explanation:
1. S = \( \frac{I_g G}{I - I_g} \) = \( \frac{0.005 \times 20}{1 - 0.005} \)
2. S = \( \frac{0.1}{0.995} \) ≈ 0.1 Ω

Teacher's Note:
1. The shunt carries 995 mA and the coil only 5 mA.
2. The shunt goes in parallel with the coil.

Answer: (b) 1980 Ω

Explanation:
1. R = \( \frac{V}{I_g} \) − G = \( \frac{10}{0.005} \) − 20
2. R = 2000 − 20 = 1980 Ω

Teacher's Note:
1. 2000 Ω is the total resistance of the voltmeter; subtract the coil's own 20 Ω.
2. The series resistor limits the current to 5 mA at 10 V.

Answer: (a) 6 divisions per mA

Explanation:
1. Current sensitivity = \( \frac{\varphi}{I} \) = \( \frac{30 \text{ divisions}}{5 \text{ mA}} \)
2. = 6 divisions per mA

Teacher's Note:
1. Sensitivity is deflection per unit current, not current per division.
2. It equals NAB/k for the coil.

Answer: (b) make the field radial so that the deflection is proportional to the current

Explanation:
1. They make the magnetic field radial, so the plane of the coil is always parallel to B and the torque is NIAB at every position.
2. Then NIAB = kφ gives φ ∝ I, so the scale is linear. The answer is (b).

Teacher's Note:
1. The soft iron core also strengthens the field, which raises the sensitivity.
2. A linear scale has equally spaced divisions.

Answer: (d) 2000 Ω

Explanation:
1. For 20 V the total resistance must be \( \frac{20}{0.005} \) = 4000 Ω, so R = 4000 − 20 = 3980 Ω.
2. Already present: 1980 Ω. Extra resistance = 3980 − 1980 = 2000 Ω

Teacher's Note:
1. Read the question carefully: it asks for the extra resistance, not the total.
2. Doubling the range doubles the total resistance of the voltmeter.