JEE Physics Moving Charges and Magnetism MCQs Set 30

Answer: (b) \( r_e\lt r_p=r_{He} \)

Explanation:
1. \( r=\dfrac{\sqrt{2mK}}{qB}\propto\dfrac{\sqrt m}{q} \) for equal K.
2. Proton: \( \sqrt{1}/1=1 \). Helium nucleus: \( \sqrt4/2=1 \). Electron: \( \sqrt{1/1836} \), much smaller. Hence \( r_e\lt r_p=r_{He} \).

Teacher's Note:
1. Write the force or field law for the situation first, then substitute SI values.
2. Check units and a limiting case before choosing the option.

Answer: (c) \( 1:\sqrt2 \)

Explanation:
1. \( r=\dfrac1B\sqrt{\dfrac{2mV}{q}} \), so \( r\propto\sqrt{m/q} \).
2. \( \dfrac{r_p}{r_\alpha}=\sqrt{\dfrac{1/1}{4/2}}=\dfrac1{\sqrt2} \).

Teacher's Note:
1. Write the force or field law for the situation first, then substitute SI values.
2. Check units and a limiting case before choosing the option.

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

Explanation:
1. \( r=\dfrac{\sqrt{2meV}}{eB}=\dfrac1B\sqrt{\dfrac{2mV}{e}} \).
2. \( r=\dfrac1{0.1}\sqrt{\dfrac{2\times9.1\times10^{-31}\times500}{1.6\times10^{-19}}}=7.5\times10^{-4} \) m.

Teacher's Note:
1. Write the force or field law for the situation first, then substitute SI values.
2. Check units and a limiting case before choosing the option.

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

Explanation:
1. Straight line: \( qE=qvB \Rightarrow v=E/B=200 \) m/s.
2. Circle: \( m=\dfrac{qBr}{v}=\dfrac{1.6\times10^{-19}\times0.5\times5\times10^{-3}}{200}=2.0\times10^{-24} \) kg.

Teacher's Note:
1. Write the force or field law for the situation first, then substitute SI values.
2. Check units and a limiting case before choosing the option.

Answer: (c) \( r_e\lt r_p=r_\alpha \)

Explanation:
1. \( r\propto\sqrt m/q \) for equal kinetic energy.
2. Proton: 1, alpha: \( \sqrt4/2=1 \), electron: much smaller, so \( r_e\lt r_p=r_\alpha \).

Teacher's Note:
1. Write the force or field law for the situation first, then substitute SI values.
2. Check units and a limiting case before choosing the option.

Answer: (a) \( \vec E=a\hat i;\ \vec B=c\hat k+a\hat i \)

Explanation:
1. Motion must stay in the xy plane, so B must have no y-component, and the path is non-circular only if E changes the speed.
2. \( \vec E=a\hat i \) changes the speed along x. The z-component of B gives \( q\vec v\times\vec B=-qvc\,\hat j \), curving the path towards \( -y \). The \( \hat i \) component of B exerts no force. A y-component of B would push the particle out of the plane, and E = 0 gives a circle.

Teacher's Note:
1. Write the force or field law for the situation first, then substitute SI values.
2. Check units and a limiting case before choosing the option.

Answer: (b) all ions deflect towards the -y direction

Explanation:
1. Positive ions accelerate along +x, and \( q\vec v\times\vec B \) with \( \hat i\times\hat k=-\hat j \) pushes them to \( -y \).
2. Negative ions accelerate along \( -x \); with negative q and \( \vec v=-v\hat i \) the force is again along \( -\hat j \). Both deflect towards \( -y \).

Teacher's Note:
1. Write the force or field law for the situation first, then substitute SI values.
2. Check units and a limiting case before choosing the option.

Answer: (d) \( r_\alpha=r_p\lt r_d \)

Explanation:
1. \( r\propto\sqrt m/q \): proton 1, deuteron \( \sqrt2/1=\sqrt2 \), alpha \( \sqrt4/2=1 \).
2. So \( r_\alpha=r_p\lt r_d \).

Teacher's Note:
1. Write the force or field law for the situation first, then substitute SI values.
2. Check units and a limiting case before choosing the option.

Answer: (a) \( (R_1/R_2)^2 \)

Explanation:
1. \( r=\dfrac1B\sqrt{\dfrac{2mV}{q}} \Rightarrow m=\dfrac{qB^2r^2}{2V} \).
2. For equal q, V, B: \( m_X/m_Y=(R_1/R_2)^2 \).

Teacher's Note:
1. Write the force or field law for the situation first, then substitute SI values.
2. Check units and a limiting case before choosing the option.

Answer: (b) (i) and (iii)

Explanation:
1. The speed is 4 m/s both times and the velocity turns from the x-axis by \( 30^\circ=\pi/6 \) towards \( +y \). For a positive charge \( \hat i\times\vec B\) must point along \( +\hat j \), so \( \vec B \) is along \( -z \).
2. Time \( t=\dfrac{\pi/6}{\omega}=\dfrac{\pi M}{6QB}=10^{-2} \) s gives \( B=\dfrac{100\pi M}{6Q}=\dfrac{50\pi M}{3Q} \). So (i) and (iii).

Teacher's Note:
1. Write the force or field law for the situation first, then substitute SI values.
2. Check units and a limiting case before choosing the option.

Answer: (d) (ii) and (iv)

Explanation:
1. Each particle describes a semicircle and leaves moving opposite to its entry velocity, so the exit paths are parallel (ii).
2. The time is half the period \( \pi m/(qB) \), which depends on mass; the electron and proton exit at different times (iv).

Teacher's Note:
1. Write the force or field law for the situation first, then substitute SI values.
2. Check units and a limiting case before choosing the option.

Answer: (a) (i), (iii) and (iv)

Explanation:
1. Entry normal to the boundary means the particle reaches region III only if the radius \( r=\dfrac{mv}{qB}\gt l \), i.e. \( v\gt \dfrac{qlB}m \) (i).
2. For \( r=l \) the path is a full semicircle of length \( \pi l \), the largest possible; for \( r\gt l \) it is shorter (iii). A return trip is half a period, \( \pi m/qB \), independent of v (iv).

Teacher's Note:
1. Write the force or field law for the situation first, then substitute SI values.
2. Check units and a limiting case before choosing the option.

Answer: (c) (i), (ii) and (iv)

Explanation:
1. Work-energy theorem (the magnetic force does no work): \( qE(2a)=\tfrac12m(4v^2-v^2)=\tfrac32mv^2 \Rightarrow E=\dfrac{3mv^2}{4qa} \) (i).
2. At P the force \( qE\hat i \) is along \( \vec v \): power \( =qEv=\dfrac{3mv^3}{4a} \) (ii). At Q the velocity is along \( -\hat j \), perpendicular to E and to the magnetic force, so both powers are zero (iv).

Teacher's Note:
1. Write the force or field law for the situation first, then substitute SI values.
2. Check units and a limiting case before choosing the option.

Answer: (d) \( E\neq0,\ B=0 \)

Explanation:
1. With only E present the force qE changes the velocity.
2. E = 0 (with B = 0, or B parallel to v), and \( E\neq0,B\neq0 \) with \( qE=-q\vec v\times\vec B \) all allow constant velocity; \( E\neq0,B=0 \) does not.

Teacher's Note:
1. Write the force or field law for the situation first, then substitute SI values.
2. Check units and a limiting case before choosing the option.

Answer: (a) moves in a straight line with constant velocity

Explanation:
1. Net force \( =-e(\vec E+\vec v\times\vec B)=-e\left(-E_0\hat x+\dfrac{E_0}{B_0}B_0(\hat y\times\hat z)\right) \).
2. \( \hat y\times\hat z=\hat x \), so the net force is \( -e(-E_0+E_0)\hat x=0 \): constant velocity along a straight line.

Teacher's Note:
1. Write the force or field law for the situation first, then substitute SI values.
2. Check units and a limiting case before choosing the option.

Answer: (c) a helix about the z-axis with increasing pitch

Explanation:
1. The velocity component along x is perpendicular to B, giving circular motion in the xy plane.
2. E along z accelerates the particle along B, so the circle is stretched into a helix about z whose pitch increases with time.

Teacher's Note:
1. Write the force or field law for the situation first, then substitute SI values.
2. Check units and a limiting case before choosing the option.

Answer: (d) a much smaller radius, about 1/43 of that of the proton

Explanation:
1. \( r=\dfrac{\sqrt{2mK}}{qB}\propto\sqrt m \) for equal K and equal \( |q| \).
2. \( \dfrac{r_e}{r_p}=\sqrt{\dfrac{m_e}{m_p}}=\sqrt{\dfrac1{1836}}\approx\dfrac1{43} \).

Teacher's Note:
1. Write the force or field law for the situation first, then substitute SI values.
2. Check units and a limiting case before choosing the option.

Answer: (b) the magnetic force is always perpendicular to the velocity and does no work

Explanation:
1. \( \vec F=q\vec v\times\vec B \) is perpendicular to \( \vec v \).
2. Power \( =\vec F\cdot\vec v=0 \), so the kinetic energy is constant.

Teacher's Note:
1. Write the force or field law for the situation first, then substitute SI values.
2. Check units and a limiting case before choosing the option.

Answer: (c) \( \dfrac{mv_0}{2B_0q} \)

Explanation:
1. Radius \( R=\dfrac{mv_0}{qB_0} \). Turning through \( 30^\circ \) over a width L: \( L=R\sin30^\circ \).
2. \( L=\dfrac{R}{2}=\dfrac{mv_0}{2qB_0} \).

Teacher's Note:
1. Write the force or field law for the situation first, then substitute SI values.
2. Check units and a limiting case before choosing the option.

Answer: (d) \( -v_0\hat i,\ \dfrac{\pi m}{B_0q} \)

Explanation:
1. The radius is \( R=2L \) and the width is \( 2.1L\gt R \), so the particle completes a semicircle and returns to x = 0.
2. The exit velocity is \( -v_0\hat i \) and the time is half a period, \( \dfrac{\pi m}{qB_0} \).

Teacher's Note:
1. Write the force or field law for the situation first, then substitute SI values.
2. Check units and a limiting case before choosing the option.

Answer: (b) \( 1.2\times10^{-2}\ \text m,\ 4.37\times10^{-2}\ \text m \)

Explanation:
1. \( r=\dfrac{mv\sin60^\circ}{qB}=\dfrac{1.67\times10^{-27}\times4\times10^5\times0.866}{1.6\times10^{-19}\times0.3}=1.2\times10^{-2} \) m.
2. Pitch \( =v\cos60^\circ\cdot\dfrac{2\pi m}{qB}=2\times10^5\times\dfrac{2\pi\times1.67\times10^{-27}}{4.8\times10^{-20}}=4.37\times10^{-2} \) m.

Teacher's Note:
1. Write the force or field law for the situation first, then substitute SI values.
2. Check units and a limiting case before choosing the option.

Answer: (c) \( (6.4\ \text m,\ 0,\ 0) \)

Explanation:
1. Magnetic force \( qvB=1.6\times10^{-19}\times1.28\times10^6\times0.08=1.64\times10^{-14} \) N along \( +z \) (\( \hat i\times\hat j=\hat k \)); electric force \( q|E_z|=1.64\times10^{-14} \) N along \( -z \).
2. They cancel, so the particle moves undeflected: \( x=vt=1.28\times10^6\times5\times10^{-6}=6.4 \) m.

Teacher's Note:
1. Write the force or field law for the situation first, then substitute SI values.
2. Check units and a limiting case before choosing the option.

Answer: (a) 0.1 T

Explanation:
1. \( E=\dfrac Vd=\dfrac{600}{3\times10^{-3}}=2\times10^5 \) V/m.
2. Undeflected: \( eE=evB \Rightarrow B=\dfrac Ev=\dfrac{2\times10^5}{2\times10^6}=0.1 \) T.

Teacher's Note:
1. Write the force or field law for the situation first, then substitute SI values.
2. Check units and a limiting case before choosing the option.

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

Explanation:
1. Perpendicular distance of P from AB: \( d=\sqrt{5^2-3^2}=4 \) cm; \( \sin\alpha=\sin\beta=\dfrac35 \).
2. \( B=\dfrac{\mu_0I}{4\pi d}(\sin\alpha+\sin\beta)=\dfrac{10^{-7}\times5}{0.04}\times1.2=1.5\times10^{-5} \) T.

Teacher's Note:
1. Write the Biot-Savart or Ampere result for each element of the geometry and add the fields as vectors.
2. Check that the answer reduces correctly in a limiting case such as zero current or a straight infinite wire.

Answer: (c) \( 3\times10^{-5}\ \text C \)

Explanation:
1. Equivalent current \( I=\dfrac{q\omega}{2\pi} \) and \( B=\dfrac{\mu_0I}{2R}=\dfrac{\mu_0q\omega}{4\pi R} \).
2. \( q=\dfrac{4\pi RB}{\mu_0\omega}=\dfrac{4\pi\times0.1\times3.8\times10^{-9}}{4\pi\times10^{-7}\times40\pi}\approx3\times10^{-5} \) C.

Teacher's Note:
1. Write the Biot-Savart or Ampere result for each element of the geometry and add the fields as vectors.
2. Check that the answer reduces correctly in a limiting case such as zero current or a straight infinite wire.