JEE Physics Moving Charges and Magnetism MCQs Set 21

Answer: (a) constant in magnitude only

Explanation:
1. By the Biot-Savart law for a moving charge, \( B=\dfrac{\mu_0}{4\pi}\dfrac{q\,\vec v\times\vec r}{r^3} \).
2. Here \( v \) is perpendicular to \( \vec r \) and \( r \) is fixed, so the magnitude is constant.
3. The direction of \( \vec v\times\vec r \) turns around the axis as the charge moves, so the direction changes.

Teacher's Note:
1. Check separately whether the magnitude and the direction of a vector are constant.
2. A uniform circular motion keeps speed fixed but not velocity.

Answer: (b) 2

Explanation:
1. At the midpoint both wires are at the same distance, so the field is proportional to the currents.
2. Same direction: \( i_1-i_2=10k \). Opposite direction: \( i_1+i_2=30k \).
3. Adding and subtracting: \( i_1=20k,\ i_2=10k \), so the ratio is 2.

Teacher's Note:
1. Between parallel currents the fields subtract; between antiparallel currents they add.
2. Use the two readings as two simultaneous equations.

Answer: (a) \( \dfrac{\mu_0 I}{12\pi} \)

Explanation:
1. Wire along x is 3 m from P: \( B_1=\dfrac{\mu_0I}{2\pi\cdot3} \), directed out of the page (current +x, point above the wire).
2. Wire along y is 2 m from P: \( B_2=\dfrac{\mu_0I}{2\pi\cdot2} \), directed into the page (current +y, point to its right).
3. Net \( =\dfrac{\mu_0I}{2\pi}\left(\dfrac12-\dfrac13\right)=\dfrac{\mu_0I}{12\pi} \).

Teacher's Note:
1. Use the right-hand thumb rule for each wire and then subtract the opposite fields.
2. Always find the direction before adding the magnitudes.

Answer: (c) 4/3

Explanation:
1. For a long wire \( B\propto 1/r \).
2. So \( r_1/r_2=B_2/B_1=4/3 \).

Teacher's Note:
1. The field of a long straight wire falls off as the first power of the distance.
2. Inverse proportion inverts the ratio.

Answer: (c) \( \dfrac{\mu_0 i}{2\pi}\left(\dfrac1{r-x}+\dfrac1x\right) \)

Explanation:
1. Between two oppositely directed currents the fields of both wires point the same way.
2. So \( B=\dfrac{\mu_0 i}{2\pi x}+\dfrac{\mu_0 i}{2\pi(r-x)} \).

Teacher's Note:
1. For antiparallel currents the field is strongest in the region between the wires.
2. For parallel currents the fields cancel at the midpoint.

Answer: (d) 2500 turns/m

Explanation:
1. \( B=\mu_0 n I\Rightarrow n=\dfrac{3.14\times10^{-2}}{4\pi\times10^{-7}\times10} \).
2. Evaluating, \( n=2500 \) turns/m.

Teacher's Note:
1. Inside a long solenoid the field depends only on turns per unit length and current.
2. Do not divide by the total number of turns.

Answer: (b) End A becomes positively charged

Explanation:
1. The force on a free positive charge is \( q\vec v\times\vec B \). With \( \vec v=\hat x \) and \( \vec B=-\hat z \) this is along \( +\hat y \) (upwards).
2. Positive charge collects at the top end A and the bottom end B is left negative, until the electric force balances the magnetic force.
3. No steady current flows in an open rod, so there is no Joule heating.

Teacher's Note:
1. A moving conductor in a magnetic field is a source of motional emf.
2. Use the cross product carefully for direction.

Answer: (d) \( \mu_0\lambda \)

Explanation:
1. The cylinder behaves like an infinite solenoid with \( nI=\lambda \).
2. Hence \( B=\mu_0\lambda \) inside, uniform, and zero outside.

Teacher's Note:
1. A surface current sheet wrapped around a cylinder is a solenoid.
2. Field outside an ideal solenoid is zero.

Answer: (a) \( \dfrac{\sqrt5}{2\sqrt2}\,4\pi\times10^{-5}\ \text{T} \)

Explanation:
1. Coil A: \( B_A=\dfrac{\mu_0 I_A}{2r_A}\approx 8.89\times10^{-5}\ \text{T} \).
2. Coil B: \( B_B=\dfrac{\mu_0 I_B}{2r_B}\approx 4.44\times10^{-5}\ \text{T} \).
3. The planes are perpendicular, so the fields are perpendicular: \( B=\sqrt{B_A^2+B_B^2}\approx9.93\times10^{-5}\ \text{T} \), which equals \( \dfrac{\sqrt5}{2\sqrt2}\,4\pi\times10^{-5} \) T.

Teacher's Note:
1. Fields of coils with perpendicular planes add as perpendicular vectors.
2. Compute each field with \( \mu_0I/2r \) first.

Answer: (b) 12 cm

Explanation:
1. \( qV=\tfrac12mv^2\Rightarrow \dfrac mq=\dfrac{2V}{v^2}=\dfrac{2\times12000}{10^{12}}=2.4\times10^{-8} \) kg/C.
2. Radius \( r=\dfrac{mv}{qB}=\dfrac{2.4\times10^{-8}\times10^6}{0.2}=0.12 \) m = 12 cm.

Teacher's Note:
1. The radius depends on the specific charge, so eliminate it using the accelerating potential.
2. Convert metres to centimetres at the end.

Answer: (d) a helix

Explanation:
1. The velocity component along \( \vec B \) (\( 5\hat i \)) is unaffected and gives uniform motion along the field.
2. The perpendicular component (\( 4\hat j \)) gives circular motion.
3. Together they give a helical path.

Teacher's Note:
1. Both a parallel and a perpendicular component of velocity are present.
2. Parallel component: translation; perpendicular component: rotation.

Answer: (c) \( He^+ \) and \( O^{2+} \) will be deflected equally

Explanation:
1. \( r=\dfrac{\sqrt{2mK}}{qB} \), so deflection (curvature) is proportional to \( q/\sqrt m \).
2. In units of \( e/\sqrt{m_H} \): H+ gives 1, He+ gives \( 1/\sqrt4=0.5 \), O2+ gives \( 2/\sqrt{16}=0.5 \).
3. So He+ and O2+ bend equally and H+ bends the most.

Teacher's Note:
1. With equal kinetic energy compare \( q/\sqrt m \).
2. Mass numbers 1, 4 and 16 were used.

Answer: (d) \( \sqrt3\,\pi \)

Explanation:
1. Period \( T=\dfrac{2\pi m}{qB}=2\pi \) s.
2. Pitch \( p=v\cos30^\circ\cdot T=\dfrac{\sqrt3}{2}\times2\pi=\sqrt3\pi \) m.

Teacher's Note:
1. Pitch uses the component of velocity along the field.
2. Do not use the perpendicular component.

Answer: (b) \( \dfrac{2\mu_0 I i}{3\pi} \)

Explanation:
1. Force on the near side (distance \( L/2 \)): \( F_1=\dfrac{\mu_0 I i}{2\pi (L/2)}L=\dfrac{\mu_0Ii}{\pi} \), attractive.
2. Force on the far side (distance \( 3L/2 \)): \( F_2=\dfrac{\mu_0Ii}{2\pi(3L/2)}L=\dfrac{\mu_0Ii}{3\pi} \), repulsive.
3. The forces on the two sides perpendicular to the wire cancel. Net \( F=F_1-F_2=\dfrac{2\mu_0Ii}{3\pi} \).

Teacher's Note:
1. Parallel currents attract and antiparallel currents repel.
2. The length L cancels because the force per unit length goes as 1/distance.

Answer: (c) the shape of the loop

Explanation:
1. \( \vec\tau=\vec M\times\vec B=NIA\,B\sin\theta \).
2. The torque depends on the area, not on how the area is shaped.

Teacher's Note:
1. Only the area enters the magnetic moment.
2. The same area in any shape gives the same torque.

Answer: (c) zero

Explanation:
1. The net magnetic moment of a toroid is zero because the moments of its turns, which point along the tangent, cancel around the ring.
2. So \( \vec\tau=\vec M\times\vec B=0 \) for every direction of the field.

Teacher's Note:
1. A closed ring winding has no net dipole moment.
2. Therefore there is no torque.

Answer: (b) \( 40\pi \) rad/s²

Explanation:
1. Magnetic moment \( M=I\pi R^2=4\pi(0.5)^2=\pi \) A m²; torque \( \tau=MB=10\pi \) N m.
2. The ring turns about a diameter: \( I_d=\tfrac12mR^2=\tfrac12(2)(0.25)=0.25 \) kg m².
3. So \( \alpha=\tau/I_d=40\pi \) rad/s².

Teacher's Note:
1. The axis of rotation is a diameter, not the axis of the ring.
2. Moment of inertia about a diameter is \( \tfrac12mR^2 \).

Answer: (b) \( 8\,nBaI\cos60^\circ \)

Explanation:
1. Magnetic moment \( M=(2n)(2I)(2a)=8naI \).
2. The plane is at 60° to the field, so the normal is at 30° to it: \( \tau=MB\sin30^\circ=MB\cos60^\circ \).
3. Hence \( \tau=8nBaI\cos60^\circ \).

Teacher's Note:
1. The angle in the torque formula is between the normal and the field.
2. The angle between the plane and the field is its complement.

Answer: (c) 27 : 1

Explanation:
1. On the equatorial line of a short magnet \( B\propto1/d^3 \).
2. So \( B_A/B_B=(3x/x)^3=27 \).

Teacher's Note:
1. A short magnet is a dipole, so the field falls as the cube of distance.
2. The nearer point has the stronger field.

Answer: (d) behaves like a paramagnetic substance

Explanation:
1. Above the Curie temperature thermal agitation destroys the domain alignment.
2. The material then shows only paramagnetic behaviour.

Teacher's Note:
1. Ferromagnetism disappears above the Curie point.
2. The susceptibility then follows Curie's law.