JEE Physics Moving Charges and Magnetism MCQs Set 40

Answer: (b) \( 3.33\times10^{-9} \) T

Explanation:
1. \( B \propto \frac1r \), so \( B' = 10^{-8}\times\frac{4}{12} = 3.33\times10^{-9} \) T.

Teacher's Note:
1. For a long straight wire B is inversely proportional to distance.
2. The distance is tripled, so B becomes one third.

Answer: (a) zero

Explanation:
1. The two arcs have lengths \( l_1, l_2 \) and resistances in that ratio, so \( i_1l_1 = i_2l_2 \).
2. The fields at the centre are \( \frac{\mu_0i_1l_1}{4\pi a^2} \) and \( \frac{\mu_0i_2l_2}{4\pi a^2} \), equal and opposite, so the net field is zero.

Teacher's Note:
1. Currents in parallel arcs divide in inverse ratio of arc lengths.
2. The two contributions have opposite directions.

Answer: (c) zero

Explanation:
1. By Ampere's law the net current enclosed by a circle outside the cable is i − i = 0.
2. So B = 0 outside.

Teacher's Note:
1. Equal and opposite currents give no field outside.
2. This is why coaxial cables do not radiate magnetic fields.

Answer: (c) \( 25.1\times10^{-3} \) T, \( 12.6\times10^{-3} \) T

Explanation:
1. \( n = 5000 \) turns per metre. Inside: \( B = \mu_0nI = 4\pi\times10^{-7}\times5000\times4 = 25.1\times10^{-3} \) T.
2. At the end of a long solenoid the field is half of this, 12.6 × 10⁻³ T.

Teacher's Note:
1. Convert turns per cm to turns per metre.
2. End-field is half of the central field.

Answer: (d) \( \frac{5\mu_0i}{36\pi R} \)

Explanation:
1. Current enclosed \( = i\frac{(3R/2)^2-R^2}{(2R)^2-R^2} = i\cdot\frac{5/4}{3} = \frac{5i}{12} \).
2. \( B = \frac{\mu_0I_{enc}}{2\pi r} = \frac{\mu_0\cdot5i/12}{2\pi\cdot3R/2} = \frac{5\mu_0i}{36\pi R} \).

Teacher's Note:
1. Only the part of the current inside the Amperian loop counts.
2. Current is proportional to the area of the ring enclosed.

Answer: (b) The α-particle and the proton will be bent in circular paths of the same radius

Explanation:
1. \( r = \frac{\sqrt{2mK}}{qB} \).
2. Proton: \( \frac{\sqrt{2mK}}{eB} \). α-particle: \( \frac{\sqrt{2(4m)K}}{2eB} = \frac{\sqrt{2mK}}{eB} \). Equal.

Teacher's Note:
1. Express the radius in terms of kinetic energy, not speed.
2. Mass factor √4 = 2 cancels the charge factor 2.

Answer: (c) 10 μs

Explanation:
1. Proton period \( = 25/5 = 5\ \mu s \). \( T = \frac{2\pi m}{qB} \).
2. α-particle has \( m/q \) twice that of proton, so \( T = 10\ \mu s \).

Teacher's Note:
1. Period of revolution does not depend on speed.
2. T is proportional to m/q.

Answer: (a) \( 10^{-3} \) T in the −z direction

Explanation:
1. \( B_{min} = \frac{F}{qv} = \frac{10^{-10}}{10^{-12}\times10^5} = 10^{-3} \) T, when B is perpendicular to v.
2. For a positive charge, \( \vec F = q\vec v\times\vec B \): \( \hat i\times(-\hat k) = +\hat j \), so B points along −z.

Teacher's Note:
1. Minimum B corresponds to sin θ = 1.
2. Use the right-hand rule: x̂ × (−ẑ) = ŷ.

Answer: (b) 4 : 3

Explanation:
1. \( r = \frac{mv}{qB} \propto \frac vq \).
2. \( \frac{r_1}{r_2} = \frac{2/1}{3/2} = \frac43 \).

Teacher's Note:
1. Masses are equal so they cancel.
2. Use r ∝ v/q.

Answer: (d) 3 × 10⁻² N

Explanation:
1. 500 G = 0.05 T. \( F = BIL\sin30^\circ = 0.05\times3\times0.4\times0.5 = 0.03 \) N.

Teacher's Note:
1. 1 gauss = 10⁻⁴ tesla.
2. Use the component of B perpendicular to the wire.

Answer: (a) 9 cm

Explanation:
1. Forces on C: \( \frac{\mu_0\cdot15\cdot5}{2\pi x} = \frac{\mu_0\cdot10\cdot5}{2\pi(15-x)} \).
2. \( 15(15-x) = 10x \Rightarrow x = 9 \) cm.

Teacher's Note:
1. Parallel currents attract, so the two forces on C act in opposite directions.
2. The current in C cancels on both sides.

Answer: (c) \( \sqrt2F \)

Explanation:
1. \( v = \sqrt{\frac{2eV}{m}} \) and \( F = evB \propto \sqrt V \).
2. For 2V the force is \( \sqrt2F \).

Teacher's Note:
1. Speed is proportional to √V.
2. Magnetic force is proportional to speed.

Answer: (a) \( 1.62\times10^{-7} \) s

Explanation:
1. \( T = \frac{2\pi m}{qB} = \frac{2\pi\times1.65\times10^{-27}}{1.6\times10^{-19}\times0.1} = 6.48\times10^{-7} \) s.
2. A 90° arc is a quarter revolution: \( t = T/4 = 1.62\times10^{-7} \) s.

Teacher's Note:
1. 100 mT = 0.1 T.
2. Time for an arc is the fraction of the period.

Answer: (b) \( \frac{v}{sd} \)

Explanation:
1. Radius \( r = \frac{v}{sB} \). To avoid the wall, \( r \le d \).
2. \( B \ge \frac{v}{sd} \), so the minimum is \( \frac{v}{sd} \).

Teacher's Note:
1. The particle just grazes the wall when r = d.
2. Specific charge s = q/m.

Answer: (b) \( \frac{2v_0}{B_0\alpha} \)

Explanation:
1. Force direction \( \hat i\times(-\hat k) = +\hat j \) for a positive α, so the circle's centre is at \( (0, r) \) with \( r = \frac{v_0}{\alpha B_0} \).
2. The circle meets the y-axis again at \( y = 2r = \frac{2v_0}{B_0\alpha} \).

Teacher's Note:
1. Find which way the particle curves first.
2. The diameter through the origin ends at 2r.

Answer: (c) 180°

Explanation:
1. A deviation θ by a field of length ℓ needs \( \sin\theta = \ell/r \). Here \( \ell/r = 1.5 > 1 \), so the particle does not leave the field at the far side.
2. It completes a semicircle and comes back out the way it entered: deviation 180°.

Teacher's Note:
1. If ℓ exceeds r, the particle turns around inside the field.
2. The exit side is the entry side.

Answer: (d) any one of I, B, R is doubled

Explanation:
1. Balancing forces on a small element: \( T = BIR \).
2. T is proportional to each of B, I and R, so doubling any one of them doubles the tension.

Teacher's Note:
1. Take a small arc and balance the outward magnetic force with the tensions.
2. The relation is linear in all three quantities.

Answer: (b) 2iBL

Explanation:
1. The force on a wire of any shape in a uniform field equals i (L⃗ × B⃗), where L⃗ is the straight line joining its ends.
2. The ends are at x = 0 and x = 2L on the x-axis, so the length is 2L and F = 2iBL.

Teacher's Note:
1. Replace the curved wire by the straight chord joining its ends.
2. Force is perpendicular to the chord.

Answer: (a) the poles are cut cylindrically (concave pole pieces with a soft iron core)

Explanation:
1. Cylindrical pole pieces with a soft iron core give a radial field so the coil always lies parallel to the field.
2. That makes the deflection proportional to the current.

Teacher's Note:
1. Radial field gives a scale that is linear.
2. The coil's turns and frame affect sensitivity and damping, not the field direction.

Answer: (c) \( 4.2\times10^{-2} \) T

Explanation:
1. On the axis of a small loop, \( B = \frac{\mu_0}{4\pi}\frac{2M}{r^3} = 10^{-7}\times\frac{2\times2.1\times10^{-25}}{(10^{-10})^3} = 4.2\times10^{-2} \) T.

Teacher's Note:
1. 1 Å = 10⁻¹⁰ m.
2. Axial field of a dipole is twice the equatorial one.

Answer: (a) 5.6 A

Explanation:
1. \( \frac{\mu_0I}{2R} = 7\times10^{-5} \Rightarrow I = \frac{2\times0.05\times7\times10^{-5}}{4\pi\times10^{-7}} \approx 5.6 \) A.

Teacher's Note:
1. Convert 5 cm to metres.
2. Set the loop's field equal in magnitude to the earth's.

Answer: (d) \( \pi : 4 \)

Explanation:
1. Length L: square side \( L/4 \), area \( L^2/16 \); circle area \( \frac{L^2}{4\pi} \).
2. Ratio \( = \frac{1/16}{1/4\pi} = \frac\pi4 \).

Teacher's Note:
1. Moment = current × area.
2. For a given perimeter the circle encloses the largest area.

Answer: (b) 0.1 J

Explanation:
1. \( m = NIA = 50\times2\times\pi\times(0.04)^2 = 0.503 \) A m².
2. \( W = 2mB = 2\times0.503\times0.1 \approx 0.1 \) J.

Teacher's Note:
1. From stable (θ = 0) to unstable (θ = 180°), ΔU = 2mB.
2. Area uses radius in metres.

Answer: (c) \( \pi\mu_0f\sigma d \)

Explanation:
1. Charge \( q = \sigma\cdot2\pi a\cdot d \). Current \( I = qf = 2\pi a\sigma df \).
2. \( B = \frac{\mu_0I}{2a} = \pi\mu_0f\sigma d \).

Teacher's Note:
1. A rotating charge acts like a current q f.
2. Use the field at the centre of a circular loop.