JEE Physics Moving Charges And Magnetism MCQs Set 41

Answer: (c) \( 5.024\times10^{-3}\ \text{T} \)

Explanation:
1. n = 600/0.6 = 1000 turns/m.
2. B = μ₀ n I = 4 π x 10⁻⁷ x 1000 × 4 = 5.024 × 10⁻³ T.

Teacher's Note:
1. B = μ₀ n I.

Answer: (a) \( \vec F=q(\vec E+\vec v\times\vec B) \)

Explanation:
1. The total force is the electric force qE plus the magnetic force q(v x B).

Teacher's Note:
1. F = q(E + v x B).

Answer: (c) the deflection per unit current

Explanation:
1. Current sensitivity is θ/I = NAB/k, i.e. deflection per unit current.

Teacher's Note:
1. Deflection per unit current.

Answer: (b) \( 10^{-3}\ \Omega\text{ in parallel} \)

Explanation:
1. S = G Ig/(I - Ig) = 100 × 10⁻⁵/(1 - 10⁻⁵, about 10⁻³ ohm.
2. The shunt is connected in parallel.

Teacher's Note:
1. Shunt is a small parallel resistance.

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

Explanation:
1. The cyclotron uses a magnetic field and an electric field acting on the charge of the particle.
2. A neutral particle feels neither force, so it cannot be accelerated.

Teacher's Note:
1. Needs charge.

Answer: (d) \( 5\times10^{-5}\ \text{Wb/m}^2 \)

Explanation:
1. B = μ₀ I/(2R): for 3 A, B1 = 3 × 10⁻⁵ T; for 4 A, B2 = 4 × 10⁻⁵ T.
2. The fields are perpendicular, so B = √(3² + 4² x 10⁻⁵ = 5 × 10⁻⁵ T.

Teacher's Note:
1. Perpendicular fields add as vectors.

Answer: (c) the poles are cylindrically cut

Explanation:
1. Cylindrical pole pieces with a soft iron core give a radial field.
2. Then the torque is always the same for a given current.

Teacher's Note:
1. Cylindrical poles give a radial field.

Answer: (a) attraction when in parallel and repulsion when in series

Explanation:
1. When the wires are in parallel the currents flow in the same direction, which gives attraction.
2. When they are in series the current flows in opposite directions in the two wires, which gives repulsion.

Teacher's Note:
1. Same direction attracts.

Answer: (c) E not zero, B = 0

Explanation:
1. With E not zero and B = 0 the proton would feel the force qE and accelerate.
2. The other cases are possible, for example E perpendicular to B with v = E/B.

Teacher's Note:
1. Only E would accelerate it.

Answer: (b) \( \oint\vec B\cdot d\vec l=\mu_0 I_{enc} \)

Explanation:
1. The circulation of B around a closed loop equals μ₀ times the enclosed current.

Teacher's Note:
1. Use B, not H, with μ₀.

Answer: (c) I, II and IV

Explanation:
1. Both fields fall as 1/r², are linear in the source (charge or current element) and obey superposition.
2. The current element is a vector source, so III is false.

Teacher's Note:
1. Biot-Savart source is a vector.

Answer: (c) \( 5\times10^{-6}\ \text{T, vertically up} \)

Explanation:
1. B = μ₀ I/(2 π r) = 2 × 10⁻⁷ x 75/3 = 5 × 10⁻⁶ T.
2. For a current flowing from north to south, the right-hand rule gives the field to the east of the wire pointing upward.

Teacher's Note:
1. Use the right-hand grip rule.

Answer: (c) I, II and III

Explanation:
1. For v = E/B the forces cancel (I).
2. For v < E/B the electric force qE exceeds the magnetic force qvB, so the deflection is along E (II).
3. The direction of E is perpendicular to both v and B (III). IV is false because the deflection is along E, not perpendicular to it.

Teacher's Note:
1. Net force is along E when v is less than E/B.

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

Explanation:
1. The current does not pass through the surface bounded by the path, so the circulation is zero.
2. The field itself is not zero at points of the path, so the reason is false.

Teacher's Note:
1. Zero circulation does not mean zero field.

Answer: (c) \( 5.6\times10^{-5}\ \text{T} \)

Explanation:
1. B = mv/(er) = 9 × 10⁻³¹ x 10⁶/(1.6 × 10⁻¹⁹ x 0.1).
2. B = 5.6 × 10⁻⁵ T.

Teacher's Note:
1. r = mv/(eB).

Answer: (a) along +y

Explanation:
1. The field at the point on the +x side due to a current along +y is directed along -z (into the page).
2. F = Q (v x B) = Q v B (x x (-z)) = Q v B (+y).

Teacher's Note:
1. Use the right-hand rule.

Answer: (c) \( \vec m\times\vec B \)

Explanation:
1. The torque on a magnetic dipole is the cross product m x B.

Teacher's Note:
1. Torque = m x B.

Answer: (c) both electric and magnetic fields

Explanation:
1. Every charge produces an electric field.
2. A moving charge also constitutes a current element and produces a magnetic field.

Teacher's Note:
1. Moving charge: E and B.

Answer: (c) zero

Explanation:
1. F = q v B sin(θ) with θ = 0 gives zero.

Teacher's Note:
1. Zero when v is parallel to B.

Answer: (a) 0

Explanation:
1. A circular Amperian loop of radius r less than a encloses no current.
2. So B = 0 inside the tube.

Teacher's Note:
1. Hollow tube: no field inside.

Answer: (b) \( \vec B\perp\vec v \)

Explanation:
1. B is along v x r, so it is perpendicular to v (and to r).
2. The field varies as 1/r squared.

Teacher's Note:
1. B is perpendicular to v.

Answer: (b) \( 8\times10^{-5}\ \text{N} \)

Explanation:
1. F = μ₀ I1 I2 l/(2 π d) = 2 × 10⁻⁷ x 10 × 2 × 2/0.1.
2. F = 8 × 10⁻⁵ N (repulsive).

Teacher's Note:
1. F = 2e-7 I1 I2 l/d.

Answer: (b) 1 : 4

Explanation:
1. For one turn of radius R: B1 = μ₀ I/(2R).
2. For two turns the radius is R/2: B2 = 2 μ₀ I/(2 (R/2)) = 4 B1.
3. B1 : B2 = 1 : 4.

Teacher's Note:
1. N doubles and R halves.

Answer: (b) A-2, B-4, C-1, D-3

Explanation:
1. Torque = M B sin(θ) = i A B sin(θ) (A-2).
2. Force per unit length = μ₀ i1 i2/(2 π d) (B-4).
3. B at the centre = μ₀ i/(2R) (C-1).
4. r = mv/(qB) (D-3).

Teacher's Note:
1. Standard expressions.

Answer: (a) circular and endless

Explanation:
1. Field lines of a straight current are concentric circles about the wire.
2. Magnetic field lines are closed, so they have no ends.

Teacher's Note:
1. Circles around the wire.