JEE Physics Moving Charges and Magnetism MCQs Set 08

Answer: (c) changing the applied magnetic field until qB = 2 pi m f_osc

Explanation:
1. The oscillator works at a fixed frequency, and the matching condition \( qB=2\pi mf_{osc} \) is met by varying the magnetic field.
2. Changing the voltage amplitude changes only the energy gained per gap crossing, not the frequency match.

Teacher's Note:
1. The voltage amplitude does not alter the revolution frequency.
2. The condition means that the revolution frequency equals the oscillator frequency.

Answer: (a) p and r

Explanation:
1. In a uniform field the net force on a dipole is zero (p true).
2. Torque \( \tau=\mu B\sin90^\circ=\mu B \) (r true, q false).
3. Potential energy \( U=-\mu B\cos90^\circ=0 \) (s false).

Teacher's Note:
1. The force on a dipole in a uniform field is always zero.
2. At 90 degrees, torque is maximum and U = 0.

Answer: (c) p, q and s

Explanation:
1. Uniform field: force zero (p).
2. sin 180° = 0, so torque is zero (q); r is false.
3. Potential energy \( U=-\mu B\cos180^\circ=+\mu B \) (s true).

Teacher's Note:
1. Antiparallel orientation is an unstable equilibrium with maximum potential energy.
2. Torque is zero at both 0 and 180 degrees.

Answer: (c) 20 mm

Explanation:
1. \( v=\sqrt{\frac{2eV}{m}}=1.88\times10^7\ \text{m/s} \).
2. \( T=\frac{2\pi m}{eB}=1.23\times10^{-9}\ \text{s} \).
3. Pitch \( =v\cos30^\circ\,T=1.88\times10^7\times0.866\times1.23\times10^{-9}\approx0.020\ \text{m}=20\ \text{mm} \).

Teacher's Note:
1. Find the speed from the accelerating voltage first.
2. Pitch uses only the velocity component along B.

Answer: (c) 19.7 A

Explanation:
1. Each element feels a force \( IB\,dl \) perpendicular to the field line and tangential to the ring; the horizontal components cancel and the vertical components add.
2. Vertical component per element: \( IB\,dl\sin20^\circ \). Total: \( F=I(2\pi r)B\sin20^\circ \).
3. Setting \( F=mg \): \( I=\frac{0.03\times9.8}{2\pi\times0.014\times0.5\times0.34}\approx19.7\ \text{A} \).

Teacher's Note:
1. The magnetic force on a ring element is perpendicular to both B and dl, so its vertical component is IB dl sin(angle from vertical).
2. Horizontal components of opposite elements cancel.

Answer: (b) 1.18 N m

Explanation:
1. The plane of the coil contains the field direction, so the moment (along y) is perpendicular to B and \( \tau=NIAB \).
2. \( \tau=15\times2\times\pi(0.25)^2\times0.2=1.18\ \text{N m} \).

Teacher's Note:
1. With B in the plane of the loop the torque is maximum.
2. Area = pi r^2 with r in metres.

Answer: (b) 1

Explanation:
1. Forces on the two sides parallel to x cancel (same field along each, opposite currents, equal and opposite forces).
2. The two sides parallel to y (length l) are at x and x + l, where the fields differ by \( B_0\frac{l}{l}=B_0 \).
3. Net force \( =Il\,[B(x+l)-B(x)]=IlB_0 \), so k = 1.

Teacher's Note:
1. Only the field difference between the two opposite sides contributes to the net force.
2. The y-directed arms see fields B(x) and B(x+l).

Answer: (b) 1 T

Explanation:
1. Magnetic forces on the ring elements are radial. For half a ring, balancing the force on the half with the tension at the cut: \( 2T=BI(2r_0) \).
2. So \( B=\frac{T}{Ir_0}=\frac{1.5}{10\times0.15}=1\ \text{T} \).

Teacher's Note:
1. For a current loop in a perpendicular field, tension T = B I r.
2. The effective length of a half ring is its diameter 2 r.

Answer: (d) \( \frac{\pi^2}{8\sqrt2} \)

Explanation:
1. Circle: \( 2\pi R=l\Rightarrow R=\frac{l}{2\pi} \), so \( B_A=\frac{\mu_0I}{2R}=\frac{\pi\mu_0I}{l} \).
2. Square of side \( a=\frac l4 \): each side at distance \( \frac a2 \) gives \( \frac{\mu_0I}{4\pi(a/2)}\cdot2\sin45^\circ=\frac{\mu_0I\sqrt2}{2\pi a} \); four sides: \( B_B=\frac{2\sqrt2\mu_0I}{\pi a}=\frac{8\sqrt2\mu_0I}{\pi l} \).
3. \( \frac{B_A}{B_B}=\frac{\pi\mu_0I/l}{8\sqrt2\mu_0I/(\pi l)}=\frac{\pi^2}{8\sqrt2} \).

Teacher's Note:
1. The wire length l fixes both the circle radius and the square side (l/4).
2. Use B = (mu0 I/4 pi d)(sin a + sin b) for each straight side.

Answer: (b) \( \sqrt2 \)

Explanation:
1. Moment \( m=I\pi R^2 \) and centre field \( B=\frac{\mu_0I}{2R} \).
2. Doubling m at constant I doubles the area, so \( R_2=\sqrt2R_1 \).
3. \( \frac{B_1}{B_2}=\frac{R_2}{R_1}=\sqrt2 \).

Teacher's Note:
1. With I fixed, m is proportional to R squared and B is proportional to 1/R.
2. So B1/B2 = sqrt(m2/m1).

Answer: (d) \( 7.5\times10^{-4}\ \text{m} \)

Explanation:
1. \( v=\sqrt{\frac{2eV}{m}}=\sqrt{\frac{2\times1.6\times10^{-19}\times500}{9.1\times10^{-31}}}=1.33\times10^7\ \text{m/s} \).
2. \( r=\frac{mv}{eB}=\frac{9.1\times10^{-31}\times1.33\times10^7}{1.6\times10^{-19}\times0.1}=7.5\times10^{-4}\ \text{m} \).

Teacher's Note:
1. Convert 100 mT to 0.1 T.
2. Equivalent shortcut: r = sqrt(2mV/e)/B.

Answer: (d) \( 2.0\times10^{-24}\ \text{kg} \)

Explanation:
1. Straight path: \( qE=qvB\Rightarrow v=\frac EB=\frac{100}{0.5}=200\ \text{m/s} \).
2. Circular path: \( m=\frac{qBr}{v}=\frac{1.6\times10^{-19}\times0.5\times5\times10^{-3}}{200}=2.0\times10^{-24}\ \text{kg} \).

Teacher's Note:
1. Velocity selector condition gives the speed.
2. Then the radius formula gives the mass.

Answer: (d) \( F\propto\left(\frac ad\right)^2 \)

Explanation:
1. By Newton's third law the force on the wire equals the force on the loop.
2. The wire's field is non-uniform: \( B=\frac{\mu_0I}{2\pi x} \). A small loop of moment \( m=i\pi a^2 \) in a field gradient feels \( F=m\left|\frac{dB}{dx}\right|=i\pi a^2\cdot\frac{\mu_0I}{2\pi d^2} \).
3. So \( F\propto\frac{a^2}{d^2}=\left(\frac ad\right)^2 \).

Teacher's Note:
1. A uniform field gives zero net force on a loop; the force here comes from the field gradient dB/dx.
2. Force on a dipole is m dB/dx along the gradient.

Answer: (a) \( T_h=T_c \)

Explanation:
1. For small oscillations \( T=2\pi\sqrt{\frac{I}{\mu B}} \).
2. Moment of inertia about the axis through the centre: hoop \( MR^2 \), solid cylinder \( \frac12MR^2 \) (ratio 2 : 1), and \( \mu_h=2\mu_c \).
3. \( \frac{T_h}{T_c}=\sqrt{\frac{I_h/\mu_h}{I_c/\mu_c}}=\sqrt{\frac{2}{2}}=1 \).

Teacher's Note:
1. Use the moments of inertia quoted above, as the problem intends.
2. Doubling both I and mu leaves the period unchanged.

Answer: (c) \( \frac\pi4n\rho_0\ell^3 \)

Explanation:
1. A charge element \( dq=\rho_0\frac{x}{\ell}dx \) at distance x rotating n times per second is a current \( n\,dq \) in a loop of area \( \pi x^2 \).
2. \( \mu=\int n\pi x^2\,dq=\frac{n\pi\rho_0}{\ell}\int_0^\ell x^3dx=\frac{\pi n\rho_0\ell^3}{4} \).

Teacher's Note:
1. Treat each charge element as a ring of current.
2. Integrate x^2 times the charge element: the integrand becomes x^3.

Answer: (a) \( 10^{-3} \)

Explanation:
1. Balance deflecting and restoring torques: \( NIAB=k\phi \).
2. \( B=\frac{k\phi}{NIA}=\frac{10^{-6}\times(\pi/180)}{175\times10^{-3}\times10^{-4}}\approx10^{-3}\ \text{T} \).

Teacher's Note:
1. Convert the angle to radians (1 degree = 0.01745 rad) and the area to square metres.
2. Radial field assumed, so torque = NIAB.

Answer: (a) \( B\pi r^2IN \)

Explanation:
1. The coil lies in the XZ plane, so its moment is along the Y-axis, perpendicular to \( \vec B \) along X.
2. \( \tau=NIAB\sin90^\circ=B\pi r^2IN \).

Teacher's Note:
1. The angle between the moment and the field is 90 degrees, so torque is maximum.
2. N turns multiply the moment by N.

Answer: (a) \( 18\ \mu\text{T} \)

Explanation:
1. Distance from the centre to each side (inradius): \( d=\frac{1}{2\sqrt3}=0.2887\ \text{m} \).
2. One side: \( B_1=\frac{\mu_0I}{4\pi d}(\sin60^\circ+\sin60^\circ)=\frac{10^{-6}}{0.2887}\times1.732=6\ \mu\text{T} \).
3. Three sides add: \( B=3\times6=18\ \mu\text{T} \).

Teacher's Note:
1. All three sides give fields in the same direction at the centre.
2. Each side subtends 60 degrees on either side of the perpendicular.

Answer: (b) repulsive and equal to \( \frac{\mu_0I_1I_2}{4\pi} \)

Explanation:
1. Nearest side (distance a, antiparallel): repulsive force \( \frac{\mu_0I_1I_2}{2\pi a}\times a=\frac{\mu_0I_1I_2}{2\pi} \).
2. Farthest side (distance 2a, parallel): attractive force \( \frac{\mu_0I_1I_2}{2\pi(2a)}\times a=\frac{\mu_0I_1I_2}{4\pi} \).
3. The forces on the other two sides cancel. Net: \( \frac{\mu_0I_1I_2}{2\pi}-\frac{\mu_0I_1I_2}{4\pi}=\frac{\mu_0I_1I_2}{4\pi} \), repulsive.

Teacher's Note:
1. The nearer side feels the larger force and its direction decides the net force.
2. Antiparallel currents repel, parallel currents attract.

Answer: (b) 0.27 N m

Explanation:
1. Area \( A=5\times2.5\ \text{cm}^2=1.25\times10^{-3}\ \text{m}^2 \).
2. Initially the normal (Y-axis) is perpendicular to B. Tilting about Z by 45° turns the normal to 45° from the X-axis, so \( \tau=NIAB\sin45^\circ=100\times3\times1.25\times10^{-3}\times1\times0.707=0.27\ \text{N m} \).

Teacher's Note:
1. Tilt axis Z lies in the coil plane, so the normal rotates in the XY plane.
2. Torque is NIAB times the sine of the angle between normal and field.

Answer: (d) \( 1.5\times10^{-5}\ \text{T} \)

Explanation:
1. For a finite wire, \( B=\frac{\mu_0I}{4\pi d}(\sin\alpha+\sin\beta) \) with \( \sin\alpha=\sin\beta=\frac35 \) (3 cm half-length over 5 cm).
2. \( B=\frac{10^{-7}\times5}{0.04}\times\frac65=1.5\times10^{-5}\ \text{T} \).

Teacher's Note:
1. The angles are measured between the perpendicular and the lines to the ends.
2. sin = half-length / distance from P to an end = 3/5.