JEE Physics Moving Charges and Magnetism MCQs Set 09

Answer: (c) \( r_e<r_p=r_{He} \)

Explanation:
1. \( r=\frac{\sqrt{2mK}}{qB} \). The electron has very small mass, so \( r_e \) is the smallest.
2. Proton: \( \frac{\sqrt{m_p}}{e} \); helium nucleus: \( \frac{\sqrt{4m_p}}{2e}=\frac{\sqrt{m_p}}{e} \), so \( r_p=r_{He} \).

Teacher's Note:
1. For the same kinetic energy, radius is proportional to sqrt(m)/q.
2. Helium: mass 4 times, charge 2 times, same as a proton.

Answer: (c) \( \text{antiparallel to }\frac{\hat i+\hat j}{\sqrt2} \)

Explanation:
1. At t = 0, z = pi/k: \( \cos(\pi)=-1 \), so \( \vec E=-E_0\frac{\hat i+\hat j}{\sqrt2} \) and the electric force is antiparallel to \( \frac{\hat i+\hat j}{\sqrt2} \).
2. The wave moves along \( -z \) (phase kz + wt), so \( \vec B\parallel\frac{\hat i-\hat j}{\sqrt2} \) with the same cosine factor (-1).
3. The magnetic force \( q\vec v\times\vec B\propto\hat k\times\left(-\frac{\hat i-\hat j}{\sqrt2}\right)=-\frac{\hat j+\hat i}{\sqrt2} \) is also antiparallel to \( \frac{\hat i+\hat j}{\sqrt2} \).

Teacher's Note:
1. Find E at the given position and time first, then the B direction from the propagation direction (E x B along the wave velocity).
2. Both electric and magnetic forces point the same way here.

Answer: (c) 0.71 mT

Explanation:
1. \( v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2\times1.6\times10^{-13}}{1.6\times10^{-27}}}=1.41\times10^7\ \text{m/s} \).
2. Force \( F=ma=1.6\times10^{-15}\ \text{N} \) and \( F=qvB \):
3. \( B=\frac{ma}{qv}=\frac{1.6\times10^{-27}\times10^{12}}{1.6\times10^{-19}\times1.41\times10^7}=7.1\times10^{-4}\ \text{T}=0.71\ \text{mT} \).

Teacher's Note:
1. 1 MeV = 1.6e-13 J.
2. The acceleration is perpendicular to the velocity, so a = qvB/m.

Answer: (d) \( \frac23 \)

Explanation:
1. Inside (\( r<a \)): \( B=\frac{\mu_0Ir}{2\pi a^2} \), at \( r=\frac a3 \): \( \frac{\mu_0I}{6\pi a} \).
2. Outside: \( B=\frac{\mu_0I}{2\pi r} \), at 2a: \( \frac{\mu_0I}{4\pi a} \).
3. Ratio \( =\frac{1/6}{1/4}=\frac23 \).

Teacher's Note:
1. Inside a solid wire B is proportional to r; outside it is proportional to 1/r.
2. Use the correct formula for each region.

Answer: (a) \( \frac{e\mu_0nIR}{2m} \)

Explanation:
1. Inside, \( B=\mu_0nI \) (axial); the electron moves on a circle that starts at the axis.
2. The circle passes through the axis, so its farthest point from the axis is at distance \( 2r \). To avoid the surface: \( 2r\le R \).
3. \( r=\frac{mv}{eB}\le\frac R2\Rightarrow v\le\frac{e\mu_0nIR}{2m} \).

Teacher's Note:
1. A circle through the axis reaches at most one diameter (2r) away from it.
2. The electron starts moving perpendicular to B, so r = mv/eB.

Answer: (d) 0.44 m

Explanation:
1. Period: \( T=\frac{2\pi m}{qB}=\frac{2\pi\times1.67\times10^{-27}}{1.6\times10^{-19}\times0.3}=2.19\times10^{-7}\ \text{s} \).
2. Speed along the field: \( v\cos60^\circ=2\times10^5\ \text{m/s} \).
3. \( l=v_\parallel\times10T=2\times10^5\times2.19\times10^{-6}=0.44\ \text{m} \).

Teacher's Note:
1. Only the component along B carries the proton across the region.
2. Time in the region = 10 periods.

Answer: (d) 4 cm

Explanation:
1. \( T=\frac{2\pi m}{qB}=2.19\times10^{-7}\ \text{s} \).
2. Pitch \( =v\cos60^\circ\times T=2\times10^5\times2.19\times10^{-7}\approx4.4\times10^{-2}\ \text{m}\approx4\ \text{cm} \).

Teacher's Note:
1. Pitch = v cos(theta) x T.
2. Use B in tesla and mass in kg.

Answer: (c) 20

Explanation:
1. Coil plane parallel to B gives maximum torque: \( \tau=NIAB \).
2. \( B=\frac{\tau}{NIA}=\frac{1.5}{500\times0.5\times3\times10^{-4}}=20\ \text{T} \).

Teacher's Note:
1. Plane parallel to the field means sin(90 degrees) = 1 for the moment-field angle.
2. Substitute in SI units.

Answer: (d) \( -30\hat i+32\hat j-9\hat k \)

Explanation:
1. \( \vec v\times\vec B=\begin{vmatrix}\hat i&\hat j&\hat k\\2&3&4\\5&3&-6\end{vmatrix}\times10^{-3}=(-18-12)\hat i-(-12-20)\hat j+(6-15)\hat k=(-30\hat i+32\hat j-9\hat k)\times10^{-3} \).
2. \( \vec F=q\,\vec v\times\vec B=10^{-6}\times10^{-3}(\dots)=(-30\hat i+32\hat j-9\hat k)\times10^{-9}\ \text{N} \).

Teacher's Note:
1. Expand the determinant carefully; note the middle term carries a minus sign.
2. Combine the powers of ten: 1e-6 x 1e-3 = 1e-9.

Answer: (a) \( \frac{2\mu_0I^2a^2}{\pi b} \)

Explanation:
1. Since \( b\gg a \), the field of the wire is nearly uniform over the loop: \( B=\frac{\mu_0I}{2\pi b} \) along the x-direction at the origin.
2. The loop's moment \( m=I(2a)^2=4Ia^2 \) is along y (normal to the XZ plane), perpendicular to B, so \( \tau=mB=4Ia^2\cdot\frac{\mu_0I}{2\pi b}=\frac{2\mu_0I^2a^2}{\pi b} \).

Teacher's Note:
1. The torque about z comes from m x B with m along y and B along x.
2. The loop side is 2a, so its area is 4a^2.

Answer: (a) \( 4.8\times10^{-19}\ \text{N} \)

Explanation:
1. The wave travels along x with E along y, so B is along z with maximum \( B_0=\frac{E_0}{c}=\frac{30}{3\times10^8}=10^{-7}\ \text{T} \).
2. The electron moves along y, perpendicular to B: \( F=evB_0=1.6\times10^{-19}\times3\times10^7\times10^{-7}=4.8\times10^{-19}\ \text{N} \).

Teacher's Note:
1. In an EM wave E0 = c B0.
2. Speed is 0.1 c = 3e7 m/s.

Answer: (c) \( \text{The quantity }\vec v\times\vec a\text{ is constant in time} \)

Explanation:
1. \( a_x=\frac kmv_y,\quad a_y=\frac kmv_x \).
2. \( \vec v\cdot\vec a=\frac{2k}{m}v_xv_y \) is not constant, and \( \vec F\cdot\vec v\ne0 \), so the kinetic energy changes and the force cannot be magnetic.
3. \( (\vec v\times\vec a)_z=v_xa_y-v_ya_x=\frac km(v_x^2-v_y^2) \). Also \( \frac{d}{dt}(v_x^2-v_y^2)=2v_xa_x-2v_ya_y=0 \), so this quantity is constant.

Teacher's Note:
1. Check whether F is perpendicular to v: it is not, so the force is not magnetic.
2. Differentiate vx^2 - vy^2 using the given equations of motion.

Answer: (d) \( \frac{qdB_0}{m} \)

Explanation:
1. The particle moves on a circle whose centre lies on the line through its starting point perpendicular to the velocity, so it reaches a distance r from the starting plane towards the screen.
2. It just fails to hit the screen when \( r=d \): \( \frac{mv}{qB_0}=d \).
3. \( v_{max}=\frac{qB_0d}{m} \).

Teacher's Note:
1. The farthest point along the original direction of motion is one radius away.
2. Larger speed means larger radius, so the condition gives a maximum speed.

Answer: (a) \( 5\times10^{-3}\ \text{T, along }+z \)

Explanation:
1. E = 300 V/cm = \( 3\times10^4 \) V/m. Electric force on the electron: \( -eE\hat j \) (along -y).
2. The magnetic force must be along +y: \( -e\vec v\times\vec B=+\hat j \) needs \( \hat i\times\vec B=-\hat j \), i.e. \( \vec B\parallel+\hat k \) because \( \hat i\times\hat k=-\hat j \).
3. Magnitude: \( B=\frac Ev=\frac{3\times10^4}{6\times10^6}=5\times10^{-3}\ \text{T} \).

Teacher's Note:
1. Convert V/cm to V/m first.
2. The two forces must be equal and opposite: eE = evB.

Answer: (c) III and IV only

Explanation:
1. E and B are both along y. The velocity component perpendicular to y circles (period constant), while the electric force accelerates the charge along y.
2. (I) false: for \( \theta=0^\circ \) there is also acceleration along y, so the path is a helix with increasing pitch, not a plane circle. (II) false, because the axial speed keeps increasing, so the pitch is not constant.
3. (III) true for \( \theta=10^\circ \) (helix with increasing pitch). (IV) true for \( \theta=90^\circ \): v is along B, no magnetic force, only electric acceleration along y.

Teacher's Note:
1. Parallel E and B give circular motion about the field plus acceleration along it.
2. If the velocity is along B the motion is straight and accelerated.

Answer: (c) 5

Explanation:
1. Use superposition: full cylinder (radius a) carrying J minus a cylinder of radius a/2 (the cavity) carrying J.
2. Full cylinder, at its surface point P: \( B_1=\frac{\mu_0J\pi a^2}{2\pi a}=\frac{\mu_0Ja}{2} \).
3. Cavity: its centre is at distance \( a+\frac a2=\frac{3a}{2} \) from P: \( B_2=\frac{\mu_0J\pi(a/2)^2}{2\pi(3a/2)}=\frac{\mu_0Ja}{12} \).
4. The two fields at P are opposite (the cavity current is effectively reversed): \( B=B_1-B_2=\mu_0Ja\left(\frac12-\frac1{12}\right)=\frac{5}{12}\mu_0aJ \), so N = 5.

Teacher's Note:
1. Replace the cavity by a reversed current of density J in the same place.
2. Outside a long cylinder, B = mu0 I/(2 pi r) as if the current were along its axis.

Answer: (d) graph (d)

Explanation:
1. For \( r<R/2 \) no current is enclosed, so B = 0.
2. For \( R/2<r<R \): \( B=\frac{\mu_0J\pi(r^2-R^2/4)}{2\pi r} \), which starts from zero at R/2 and increases in a curved (concave) way up to a maximum at R.
3. For \( r>R \): \( B=\frac{\mu_0I}{2\pi r} \) decreases as 1/r. The field is continuous, which matches graph (d).

Teacher's Note:
1. B must be zero in the hollow region and continuous everywhere.
2. In the conducting region B is proportional to (r^2 - R^2/4)/r, a curve that starts from zero.

Answer: (c) A and D only

Explanation:
1. (A) true: inside the hollow cylinder the solenoid field \( \mu_0nI \) (axial) is non-zero.
2. (B), (C) false: for R < r < 2R both the solenoid's axial field and the cylinder's tangential field \( \frac{\mu_0I}{2\pi r} \) exist, so the field is neither purely axial nor purely tangential.
3. (D) true: outside the solenoid the solenoid field vanishes but the cylinder's \( \frac{\mu_0I}{2\pi r} \) remains.

Teacher's Note:
1. Superpose the solenoid field (axial, only inside) and the straight-conductor field (tangential, outside the cylinder).
2. Check each region separately.

Answer: (c) 3

Explanation:
1. Let \( X_1=\frac{X_0}{3} \), so the other wire is at \( \frac{2X_0}{3} \). In units of \( \frac{\mu_0I}{2\pi X_0} \): field strengths are \( 3 \) and \( 1.5 \).
2. Same direction: fields oppose, \( B_1\propto3-1.5=1.5 \). Opposite directions: fields add, \( B_2\propto3+1.5=4.5 \).
3. Since \( R=\frac{mu}{qB}\propto\frac1B \): \( \frac{R_1}{R_2}=\frac{B_2}{B_1}=\frac{4.5}{1.5}=3 \).

Teacher's Note:
1. Between parallel currents the fields subtract; between antiparallel currents they add.
2. Radius is inversely proportional to the net field.

Answer: (c) current in wire 1 and wire 2 is in the direction PQ and SR respectively, and h is about 1.2a

Explanation:
1. Loop field on its axis at height h (counterclockwise seen from above, so upward): \( B_L=\frac{\mu_0Ia^2}{2(a^2+h^2)^{3/2}} \).
2. The wires' fields have an axial component only if the currents are opposite (for same-direction currents the axial components cancel and the horizontal ones add). Opposite currents PQ (up) in wire 1 and SR (down) in wire 2 give a downward axial field \( B_W=\frac{\mu_0I}{2\pi}\cdot\frac{2d}{d^2+h^2} \) at the point on the axis.
3. With d = a: \( \frac{a^2}{2(a^2+h^2)^{3/2}}=\frac{1}{\pi}\frac{a}{a^2+h^2} \). Solving numerically gives \( h\approx1.2a \) (at h = 1.2a both sides are about 0.82 in units of \( \mu_0I/2\pi a \)).

Teacher's Note:
1. Cancellation needs a downward axial component from the wires, which requires antiparallel currents.
2. Solve numerically for h: the equation is a^2/(2(a^2+h^2)^1.5) = (1/pi) a/(a^2+h^2) in unit form.

Answer: (a) \( \frac{\mu_0I}{4\pi a}\,6\left[\sqrt3-1\right] \)

Explanation:
1. The star (hexagram) has 12 equal sides, each of length \( \frac{2a}{\sqrt3} \); the outer vertices are at distance 2a and the inner vertices at \( \frac{2a}{\sqrt3} \) from the centre.
2. The perpendicular distance from the centre to each side is \( d=a \). The two ends of a side subtend \( 60^\circ \) (outer end) and \( 30^\circ \) (inner end) with this perpendicular, on the same side of its foot.
3. One side: \( B_1=\frac{\mu_0I}{4\pi a}(\sin60^\circ-\sin30^\circ)=\frac{\mu_0I}{4\pi a}\cdot\frac{\sqrt3-1}{2} \). All 12 sides give fields in the same direction: \( B=12B_1=\frac{\mu_0I}{4\pi a}\,6(\sqrt3-1) \).

Teacher's Note:
1. Every side gives a field in the same direction at the centre, so add magnitudes.
2. Use B = (mu0 I/4 pi d)(sin a + sin b) for each side.