JEE Physics Moving Charges And Magnetism MCQs Set 28

Answer: (b) n i

Explanation:
1. B = μ₀ n i.
2. It does not depend on the radius.

Teacher's Note:
1. B does not depend on radius.

Answer: (b) straight line

Explanation:
1. The particle starts at rest, so initially there is only the electric force along the field.
2. Its velocity then stays along B, so there is no magnetic force and the path is a straight line.

Teacher's Note:
1. v parallel to B gives no magnetic force.

Answer: (c) \( I(\vec l\times\vec B) \)

Explanation:
1. The force on a current element is F = I (l x B).

Teacher's Note:
1. F = I(l x B).

Answer: (b) repel each other

Explanation:
1. Antiparallel currents repel.

Teacher's Note:
1. Opposite currents repel.

Answer: (a) proportional to momentum

Explanation:
1. r = mv/(qB) = p/(qB).
2. So the radius is proportional to the momentum.

Teacher's Note:
1. r = p/(qB).

Answer: (a) the frequency of revolution of the charged particle equals the frequency of the a.c. voltage source

Explanation:
1. The particle must cross the gap at the moment the voltage reverses.
2. So the oscillator frequency must equal the cyclotron frequency qB/(2 π m).

Teacher's Note:
1. Oscillator frequency = cyclotron frequency.

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

Explanation:
1. For the same speed and charge magnitude, r is proportional to the mass.
2. The electron is much lighter, so its radius is smaller, and the reason explains this.

Teacher's Note:
1. r is proportional to m.

Answer: (b) The free electrons of the rod move towards the pivoted end

Explanation:
1. A point at distance r moves with speed w r perpendicular to the rod.
2. The force -e(v x B) on the electrons points towards the pivot for clockwise rotation and field into the page.
3. So the electrons drift towards O and the outer end becomes positive.

Teacher's Note:
1. Use F = -e(v x B).

Answer: (a) zero

Explanation:
1. The two arcs are in parallel, so i1 l1 = i2 l2.
2. Their fields at the centre are proportional to i l and act in opposite directions, so the resultant is zero.

Teacher's Note:
1. Arcs give equal and opposite fields.

Answer: (c) both inside and outside the rod

Explanation:
1. By Ampere's law the field is nonzero outside, and inside the rod too because only part of the current is enclosed.

Teacher's Note:
1. Field exists inside and outside a solid rod.

Answer: (c) 250 microtesla

Explanation:
1. B_axis = μ₀ I R² / [2 (R² + x²³/2)] and B_centre = μ₀ I/(2R).
2. B_centre/B_axis = (R² + x²³/2)/R³ = 5³/3³ = 125/27.
3. B_centre = 54 × 125/27 = 250 microtesla.

Teacher's Note:
1. Use the ratio of the two formulas.

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

Explanation:
1. T = 2 π r / v and r = mv/(qB).
2. So T = 2 π m / (qB).

Teacher's Note:
1. T = 2 π m / qB.

Answer: (a) \( \frac{\mu_0 I\theta}{4\pi r} \)

Explanation:
1. A full circle (θ = 2 π) gives μ₀ I/(2r).
2. An arc of angle θ gives the fraction θ/(2 π) of this: μ₀ I θ/(4 π r).

Teacher's Note:
1. B is proportional to the angle subtended.

Answer: (c) \( 10^{-19}\mu_0 \)

Explanation:
1. The charge is 2e = 3.2 × 10⁻¹⁹ C and the period is 2 s, so I = 1.6 × 10⁻¹⁹ A.
2. B = μ₀ I/(2r) = μ₀ × 1.6 × 10⁻¹⁹/1.6 = 10⁻¹⁹ μ₀.

Teacher's Note:
1. I = q/T.

Answer: (a) as long as there is current in the wire

Explanation:
1. Oersted found that the field exists only while the current flows.
2. It disappears when the current is switched off.

Teacher's Note:
1. Field exists only with current.

Answer: (a) \( \frac{\mu_0 qf}{2R} \)

Explanation:
1. The equivalent current is I = q f.
2. B = μ₀ I/(2R) = μ₀ q f/(2R).

Teacher's Note:
1. I = q f.

Answer: (c) \( 2.79\times10^{-4}\ \text{T} \)

Explanation:
1. B = μ₀ N I/(2R) = 4 π x 10⁻⁷ x 100 × 0.4 / (2 × 0.09).
2. B = 2.79 × 10⁻⁴ T.

Teacher's Note:
1. B = μ₀ N I / 2R.

Answer: (d) remains constant with time

Explanation:
1. The magnetic force does no work.
2. So the speed and the kinetic energy remain constant.

Teacher's Note:
1. No work, no change in KE.

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

Explanation:
1. If the magnetic moment is parallel or antiparallel to B, θ is 0 or 180° and the torque is zero, so the assertion is false.
2. The torque formula is correct.

Teacher's Note:
1. Torque can be zero.

Answer: (c) zero because F acts perpendicular to v

Explanation:
1. The magnetic force F = q(v x B) is always perpendicular to v.
2. So F . v = 0 and the work done is zero.

Teacher's Note:
1. Magnetic force does no work.

Answer: (d) the magnetic forces between them

Explanation:
1. Each wire lies in the magnetic field of the other and experiences a force.
2. For currents in the same direction the force is attractive.

Teacher's Note:
1. It is a magnetic force.

Answer: (c) four times

Explanation:
1. For x much greater than R, B = μ₀ I R²/(2 x³, which is proportional to R².
2. Doubling R makes B four times larger.

Teacher's Note:
1. B is proportional to R squared.

Answer: (a) the shape of the loop

Explanation:
1. Torque = M B sin(θ) with M = I A.
2. It depends on the area but not on the shape of the loop.

Teacher's Note:
1. Torque depends on area, not shape.

Answer: (a) the curved paths of the electron and the proton are the same (ignoring the sense of revolution)

Explanation:
1. r = p/(qB).
2. They have the same momentum and the same charge magnitude, so the radii are equal.

Teacher's Note:
1. r depends on p/q.

Answer: (d) the magnetic field at any point inside the pipe is zero

Explanation:
1. By Ampere's law a circular loop inside the pipe encloses no current.
2. So B = 0 at every point inside.

Teacher's Note:
1. No enclosed current inside.