Moving Charges and Magnetism MCQs

Force on a moving charge is **F = q v B sinθ**, where θ is the angle between v and B.

If the velocity is parallel to the magnetic field, the angle θ = 0°.

sin 0° = 0, so the magnetic force becomes zero.

Hence the magnetic force on the charge is zero.

From Biot‑Savart: d𝐵 = (μ₀/4π)·(I d𝐥 × r̂)/r².

The magnitude contains the factor 1/r², where r is the distance from the element to the point.

Thus the magnetic field varies inversely as the square of the distance.

Hence the correct answer is the square of the distance.

Inside a long solenoid the field lines are parallel and uniform, so Ampère’s law gives ∮B·dl = μ₀ I_enc.

Choosing a rectangular Amperian loop that runs inside the solenoid for length l, only the segment inside contributes: B l = μ₀ (n l) I.

Cancel l to obtain B = μ₀ n I, which matches the given answer.

Force per unit length between two long wires: F/L = μ₀ I₁ I₂ / (2π d).

Substitute I₁=3 A, I₂=4 A, d=0.06 m, μ₀=4π×10⁻⁷ T·m/A → F/L = (4π×10⁻⁷)(3×4) / (2π×0.06) = (4×10⁻⁷×12) / (0.12) = 4×10⁻⁵ N/m.

Since the currents are opposite, the force is repulsive.

4×10⁻⁵ N/m (repulsive)

Maximum force occurs when the wire is perpendicular to the field, so \(F = BIL\).

Substitute the given values: \(3\ \text{N}= B \times 5\ \text{A} \times 2\ \text{m}\).

Solve for \(B\): \(B = \frac{3}{5 \times 2}= \frac{3}{10}=0.3\ \text{T}\).

Thus the magnetic field strength is 0.3 T.

Use Ampère’s law for a circular path of radius r inside the wire: ∮B·dl = μ₀ I_enc.

The enclosed current I_enc = J·πr² = (I/πR²)·πr² = I (r²/R²).

Thus B·(2πr) = μ₀ I (r²/R²) ⇒ B = (μ₀ I /2πR²) r, showing B ∝ r.

Hence the magnetic field varies proportional to r.

Force per unit length between two long parallel conductors: F/L = μ₀ I₁ I₂ / (2π d).

For the definition of the ampere, I₁ = I₂ = 1 A and the force is set to 2×10⁻⁷ N per metre.

Insert μ₀ = 4π×10⁻⁷ H m⁻¹ and solve d = μ₀/(2π·2×10⁻⁷) = 1 m.

Thus the wires must be separated by 1 metre.

In a moving‑coil galvanometer the torque τ = N I A B sinθ, where θ is the angle between coil normal and field direction.

If the magnetic field is radial, the angle θ remains 90° for any coil rotation, so sinθ = 1 constantly.

Thus τ = N I A B is directly proportional to current I for every angular position, giving a linear scale and maximum torque throughout the motion.

Hence the scale is linear and torque is maximum for all positions.

Magnetic force on a moving charge: F = q v B sin θ.

Here θ = 90°, so sin θ = 1 and F = (1.6×10⁻¹⁹ C)(2×10⁶ m/s)(0.5 T).

Multiplying: 1.6×2 = 3.2; 3.2×0.5 = 1.6; powers of ten give 10⁻¹⁹ × 10⁶ = 10⁻¹³.

Thus F = 1.6×10⁻¹³ N, which matches the given answer.

Magnetic field at centre of a single loop: B = μ₀I/(2R) (derived from Biot‑Savart for a circular arc of 2π rad).

For N tightly wound turns, fields add linearly, giving B = N·μ₀I/(2R).

Thus the correct expression is B = μ₀ N I/(2R).

Magnetic field inside a long solenoid: B = μ₀ n I, where n = N/L = 1000/0.5 = 2000 turns m⁻¹.

Substitute μ₀ = 4π×10⁻⁷ T·m/A, I = 2 A: B = (4π×10⁻⁷)(2000)(2) = 5.03×10⁻³ T.

Thus the field equals 5.03×10⁻³ T.

Torque τ = IAB sinθ, where θ is the angle between the magnetic field and the normal to the loop.

Maximum torque occurs when sinθ = 1, i.e., when the normal is perpendicular to B.

If the normal is perpendicular to B, the plane of the loop is parallel to B.

Hence the torque is maximum when the plane of the loop is parallel to B.

Magnetic force on a charge is **q v × B**, which is always perpendicular to the instantaneous velocity vector.

Since work = ∫F·dr, the dot product of the force with the displacement is zero at every instant.

Therefore the magnetic force does no work on the particle, so the work done is always zero.

Magnetic field at centre of a circular loop: B = μ₀ I / (2R)

Substitute μ₀ = 4π×10⁻⁷ T·m/A, I = 2 A, R = 0.1 m → B = (4π×10⁻⁷ × 2)/(2×0.1)

Simplify: numerator = 8π×10⁻⁷, denominator = 0.2, so B = (8π×10⁻⁷)÷0.2 = 40π×10⁻⁷ = 4π×10⁻⁶ T

Numerically 4π×10⁻⁶ ≈ 1.26×10⁻⁵ T, which matches the given answer. 1.26x10^-5 T.

Inside a long solenoid the field is uniform, B_center ≈ μ₀nI.

At an open end the field lines spread out, giving roughly half the axial component of the interior field.

Thus B_end ≈ ½ B_center, which matches “half the field at the centre”.

Torque on a current loop is τ = μ × B, where magnetic moment μ = N I A (normal direction).

The magnitude is |τ| = μB sinθ = N I A B sinθ, with θ the angle between μ (loop normal) and B.

Thus the torque varies as the sine of the angle, giving τ = N I A B sinθ.

Centripetal force needed for circular motion is \(F_c = mv^2/r\).

Magnetic force on a moving charge is \(F_B = qvB\) (perpendicular to velocity).

Set \(F_c = F_B\): \(mv^2/r = qvB\).

Solve for r → \(r = mv/(qB)\).

For a straight current element the magnetic field is given by Biot‑Savart: d𝐵 = (μ₀/4π) I dℓ × r̂ / r².

The cross‑product is zero when the position vector r is parallel to the element (θ = 0°), i.e., on the line (axis) of the wire.

Hence no magnetic field is produced at points lying along the axis of the element.

The correct answer is along the axis of the element.

For a toroid the windings form a closed circular loop, so the net current enclosed by any Amperian loop drawn outside the toroid is zero.

Applying Ampère’s law ∮B·dl = μ₀I_enc gives B·(2πr) = μ₀·0, hence B = 0.

Thus no magnetic field exists in the empty interior hole or outside the toroid.

Maximum torque τ = N I A B sin90° = N I A B.

Substitute N=50, I=2 A, A=4×10⁻³ m², B=0.5 T → τ = 50×2×4×10⁻³×0.5 = 0.2 N·m.

Thus the correct answer is 0.2 N m.