JEE Physics Moving Charges and Magnetism MCQs Set 20

Answer: (c) 0.24 J

Explanation:
1. Magnetic moment \( \mu=NiA=100\times0.1\times\pi\times0.05^2=0.0785 \) A m\( ^2 \).
2. Starting with the plane perpendicular to the field (moment parallel to B), the coil is turned to the antiparallel position: \( W=2\mu B=2\times0.0785\times1.5=0.236 \) J.

Teacher's Note:
1. For a coil use mu = N i A and torque = mu B sin(theta).
2. Always ask which angle the formula uses: field to axis or field to plane.

Answer: (a) \( \sqrt2\,\pi\, i a B_0 \)

Explanation:
1. A point of the loop at angle \( \varphi \) has position \( (a\cos\varphi,a\sin\varphi,a) \), distance \( a\sqrt2 \) from the origin, so \( \vec B=\frac{B_0}{\sqrt2}(\cos\varphi,\sin\varphi,1) \).
2. With \( d\vec l=a(-\sin\varphi,\cos\varphi,0)d\varphi \): \( d\vec l\times\vec B=\frac{aB_0}{\sqrt2}(\cos\varphi,\sin\varphi,-1)d\varphi \).
3. Integrating over \( 0\le\varphi\le2\pi \) the x and y parts vanish, leaving \( F=\frac{i a B_0}{\sqrt2}\times2\pi=\sqrt2\,\pi\, i a B_0 \) along the z-axis.

Teacher's Note:
1. Use F = i l B sin(theta) with l the effective straight length between the ends.
2. A closed loop in a uniform field has zero net force.

Answer: (d) \( 60^\circ \)

Explanation:
1. \( \tau=\tau_{max}\sin\theta \) with \( \theta \) the angle between \( \vec\mu \) (the axis) and \( \vec B \).
2. Half of the maximum: \( \sin\theta=\frac12 \), so \( \theta=30^\circ \) between field and axis.
3. The plane of the coil is perpendicular to the axis, so the angle between the field and the plane is \( 90^\circ-30^\circ=60^\circ \).

Teacher's Note:
1. For a coil use mu = N i A and torque = mu B sin(theta).
2. Always ask which angle the formula uses: field to axis or field to plane.

Answer: (a) zero and 0.02 N m

Explanation:
1. The net force on a closed loop in a uniform field is zero.
2. The field lies in the plane of the loop, so the torque is maximum: \( \tau=iAB=10\times0.01\times0.20=0.02 \) N m.

Teacher's Note:
1. For a coil use mu = N i A and torque = mu B sin(theta).
2. Always ask which angle the formula uses: field to axis or field to plane.

Answer: (b) \( 3.14\times10^{-2} \) N m

Explanation:
1. \( \tau=NiAB\sin\theta=500\times1.0\times\pi(0.01)^2\times0.40\times\sin30^\circ \).
2. This equals \( 500\times3.14\times10^{-4}\times0.2=3.14\times10^{-2} \) N m.

Teacher's Note:
1. For a coil use mu = N i A and torque = mu B sin(theta).
2. Always ask which angle the formula uses: field to axis or field to plane.

Answer: (c) \( \frac{iBL^2}{4\pi} \)

Explanation:
1. Radius \( r=\frac{L}{2\pi} \), area \( A=\pi r^2=\frac{L^2}{4\pi} \).
2. Field in the plane: \( \tau=iAB=\frac{iBL^2}{4\pi} \).

Teacher's Note:
1. For a coil use mu = N i A and torque = mu B sin(theta).
2. Always ask which angle the formula uses: field to axis or field to plane.

Answer: (d) \( \frac{iBL^2}{18} \), smaller than for the circular loop

Explanation:
1. Sides \( x \) and \( 2x \) with perimeter \( 6x=L \), so \( x=\frac L6 \) and the area is \( 2x^2=\frac{L^2}{18} \).
2. Torque \( iBA=\frac{iBL^2}{18} \).
3. Compare \( \frac1{18}=0.056 \) with \( \frac1{4\pi}=0.080 \): the circle encloses the largest area for a given length, so its torque is larger.

Teacher's Note:
1. For a coil use mu = N i A and torque = mu B sin(theta).
2. Always ask which angle the formula uses: field to axis or field to plane.

Answer: (a) \( \frac{Q\omega R^2}{2} \)

Explanation:
1. Equivalent current \( i=\frac{Q}{T}=\frac{Q\omega}{2\pi} \).
2. Moment \( \mu=iA=\frac{Q\omega}{2\pi}\pi R^2=\frac{Q\omega R^2}{2} \).

Teacher's Note:
1. For a coil use mu = N i A and torque = mu B sin(theta).
2. Always ask which angle the formula uses: field to axis or field to plane.

Answer: (b) 0.111 A

Explanation:
1. The vertical coil produces a horizontal field \( B=\frac{\mu_0NI}{2r} \).
2. Set equal to the horizontal component: \( \frac{\mu_0\times100\,I}{2\times0.20}=27.8\mu_0 \).
3. So \( I=\frac{27.8\times0.40}{100}=0.111 \) A.

Teacher's Note:
1. For a coil use mu = N i A and torque = mu B sin(theta).
2. Always ask which angle the formula uses: field to axis or field to plane.

Answer: (c) 0.096 A

Explanation:
1. Vertical component \( B_V=B_H\tan30^\circ=27.8\mu_0\times0.5774 \).
2. The horizontal coil gives a vertical field \( \frac{\mu_0\times100\,I}{2\times0.30} \).
3. Equating: \( I=\frac{27.8\times0.5774\times0.60}{100}=0.0963 \) A.

Teacher's Note:
1. For a coil use mu = N i A and torque = mu B sin(theta).
2. Always ask which angle the formula uses: field to axis or field to plane.

Answer: (a) 2 V

Explanation:
1. Wire length \( L=50\times\pi\times0.20=31.4 \) m, cross-section \( \pi(10^{-4})^2=3.14\times10^{-8} \) m\( ^2 \), so \( R=\frac{2\times10^{-8}\times31.4}{3.14\times10^{-8}}=20 \) ohm (and the current from 20 V is 1 A).
2. Needed field \( \frac{\mu_0NI}{2r}=3.14\times10^{-5} \) gives \( I=\frac{3.14\times10^{-5}\times0.20}{4\pi\times10^{-7}\times50}=0.10 \) A.
3. So \( V=IR=0.10\times20=2 \) V, with the coil plane vertical and perpendicular to the horizontal field direction so that its axis lies along it.

Teacher's Note:
1. For a coil use mu = N i A and torque = mu B sin(theta).
2. Always ask which angle the formula uses: field to axis or field to plane.

Answer: (b) \( 7.5\times10^{-4} \) T m/A

Explanation:
1. Flux density \( B=\frac\Phi A=\frac{2.4\times10^{-5}}{2\times10^{-5}}=1.2 \) T (area \( 0.2\ \text{cm}^2=2\times10^{-5}\ \text{m}^2 \)).
2. Permeability \( \mu=\frac BH=\frac{1.2}{1600}=7.5\times10^{-4} \) T m/A.

Teacher's Note:
1. For a coil use mu = N i A and torque = mu B sin(theta).
2. Always ask which angle the formula uses: field to axis or field to plane.

Answer: (a) p, q and r only

Explanation:
1. For a semi-infinite wire starting at the origin with P on the z-axis, \( B=\frac{\mu_0 i}{4\pi d} \) along \( \hat u\times\hat z \), where \( \hat u \) is the direction of the wire away from O (reverse the sign for current towards O).
2. In units of \( \frac{\mu_0 i}{4\pi d} \) the fields are: p \( (0,-\tfrac43) \), q \( (0,-1.71) \), r \( (0,-1) \), s \( (-0.71,-1) \).
3. Only s has an x-component, so it is zero in p, q and r.

Teacher's Note:
1. Sketch the geometry and decide the direction of each contribution before putting in numbers.
2. Straight-wire formulas depend on the angles the ends subtend at the point.

Answer: (b) all four arrangements

Explanation:
1. Every wire lies in the xy-plane and the foot of the perpendicular from P to each wire is the origin, so each wire gives a field along \( \hat u\times\hat z \), which lies in the xy-plane.
2. So the z-component vanishes for every arrangement p, q, r and s.

Teacher's Note:
1. Sketch the geometry and decide the direction of each contribution before putting in numbers.
2. Straight-wire formulas depend on the angles the ends subtend at the point.

Answer: (c) r only

Explanation:
1. In units of \( \frac{\mu_0i}{4\pi d} \) the magnitudes are: p \( \tfrac43=1.33 \), q \( 1+\cos45^\circ=1.71 \), r \( 1 \), s \( \sqrt{0.71^2+1^2}=1.22 \).
2. Only r gives exactly 1.

Teacher's Note:
1. Sketch the geometry and decide the direction of each contribution before putting in numbers.
2. Straight-wire formulas depend on the angles the ends subtend at the point.

Answer: (d) q

Explanation:
1. Magnitudes in units of \( \frac{\mu_0i}{4\pi d} \): p 1.33, q 1.71, r 1.00, s 1.22.
2. The largest is q.

Teacher's Note:
1. Sketch the geometry and decide the direction of each contribution before putting in numbers.
2. Straight-wire formulas depend on the angles the ends subtend at the point.

Answer: (a) 1, 2, 3 and 4

Explanation:
1. In a uniform field the force on a dipole is always zero.
2. At zero angle \( \tau=\mu B\sin0=0 \), and \( U=-\mu B\cos0=-\mu B \), of magnitude \( \mu B \).
3. This is the minimum-energy orientation, so the equilibrium is stable. All four statements are true.

Teacher's Note:
1. For a coil use mu = N i A and torque = mu B sin(theta).
2. Always ask which angle the formula uses: field to axis or field to plane.

Answer: (c) 1 and 3 only

Explanation:
1. Force is zero in a uniform field.
2. The torque \( \mu B\sin\theta \) is not zero for an acute angle, and the dipole is not in equilibrium, so statements 2 and 4 are false.
3. Potential energy \( -\mu B\cos\theta \) is negative for an acute angle, so statement 3 is true.

Teacher's Note:
1. For a coil use mu = N i A and torque = mu B sin(theta).
2. Always ask which angle the formula uses: field to axis or field to plane.

Answer: (a) cycloid

Explanation:
1. The fields are \( \vec B=B_x\hat i \) and \( \vec E=E_y\hat j \), so E is perpendicular to B.
2. A particle released from rest in perpendicular E and B fields drifts at \( E/B \) while circling, which traces a cycloid.

Teacher's Note:
1. Take the charge and mass ratio first; every circular-motion result follows from it.
2. Check the direction with the right-hand rule for the sign of the charge.

Answer: (c) straight line

Explanation:
1. \( \vec u\times\vec B=0 \) means no magnetic force at the start.
2. Since \( \vec E \) is also parallel to \( \vec u \) (and so to B), the electric force only changes the speed along the same line.
3. The particle moves in a straight line (with changing speed).

Teacher's Note:
1. Take the charge and mass ratio first; every circular-motion result follows from it.
2. Check the direction with the right-hand rule for the sign of the charge.

Answer: (d) helix with variable pitch and constant radius

Explanation:
1. Perpendicular to B the velocity gives uniform circular motion of constant radius \( \frac{mu}{qB} \).
2. The electric field is along B, so it accelerates the particle along B, increasing the pitch continuously.
3. The path is a helix of constant radius and variable pitch.

Teacher's Note:
1. Take the charge and mass ratio first; every circular-motion result follows from it.
2. Check the direction with the right-hand rule for the sign of the charge.