JEE Physics Moving Charges and Magnetism MCQs Set 26

Answer: (d) \( d\vec B=\frac{\mu_0}{4\pi}\,i\left(\frac{d\vec l\times\vec r}{r^3}\right) \)

Explanation:
1. The Biot-Savart law states \( d\vec B=\frac{\mu_0}{4\pi}\frac{i\,d\vec l\times\vec r}{r^3} \).
2. Magnitude: \( dB=\frac{\mu_0}{4\pi}\frac{i\,dl\sin\theta}{r^2} \), so the cross product carries one factor of r in its length and the denominator needs \( r^3 \).

Teacher's Note:
1. The field is linear in current (power 1), not i squared.
2. Because \( |\vec r| \) appears in the cross product, the denominator is r cubed.

Answer: (b) \( 10^{-19}\,\mu_0 \)

Explanation:
1. Charge of a helium nucleus: \( q=2e=3.2\times10^{-19}\ \text{C} \); period T = 2 s.
2. Equivalent current: \( I=\frac{q}{T}=1.6\times10^{-19}\ \text{A} \).
3. \( B=\frac{\mu_0 I}{2R}=\frac{\mu_0\times1.6\times10^{-19}}{2\times0.8}=10^{-19}\mu_0 \) (in tesla).

Teacher's Note:
1. A revolving charge is equivalent to a current q/T.
2. Use the circular-loop formula B = mu0 I/2R at the centre.

Answer: (a) \( \frac{\mu_0 i}{4}\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \)

Explanation:
1. The straight segments lie along lines through O, so they contribute nothing.
2. A semicircle gives \( \frac{\mu_0 i}{4R} \). The current runs clockwise in the inner arc and anticlockwise in the outer arc, so the two fields are opposite.
3. \( B=\frac{\mu_0 i}{4R_1}-\frac{\mu_0 i}{4R_2}=\frac{\mu_0 i}{4}\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \).

Teacher's Note:
1. Check the sense of rotation of each arc before adding fields; opposite senses subtract.
2. Straight leads that point at the centre contribute zero because dl x r = 0.

Answer: (d) \( \text{zero for all values of }\theta \)

Explanation:
1. The two arcs are in parallel, so the currents are inversely proportional to their lengths: \( i_1 l_1=i_2 l_2 \).
2. Field of an arc at the centre is proportional to \( i\,l/r^2 \), so the two arcs give equal and opposite fields (the currents flow in opposite senses round the centre).
3. Net field at the centre is zero for every value of \( \theta \).

Teacher's Note:
1. Parallel arcs carry current inversely to their lengths, so i x (arc length) is the same for both.
2. That equality makes the two contributions cancel at the centre for any angle.

Answer: (b) only outside the pipe

Explanation:
1. Apply Ampere's law to a circle coaxial with the pipe and inside the hollow part: no current is enclosed, so B = 0.
2. Outside, the enclosed current is I, so \( B=\frac{\mu_0 I}{2\pi r} \).

Teacher's Note:
1. A hollow conductor behaves like a solid one for points outside it.
2. Inside the cavity the enclosed current is zero, so the field vanishes.

Answer: (c) \( 25.1\times10^{-3}\text{ Wb/m}^2,\ 12.6\times10^{-3}\text{ Wb/m}^2 \)

Explanation:
1. \( n=50\text{ per cm}=5000\text{ per m} \).
2. Inside: \( B=\mu_0 nI=4\pi\times10^{-7}\times5000\times4=25.1\times10^{-3}\ \text{T} \).
3. At one end the field is half of this: \( 12.6\times10^{-3}\ \text{T} \).

Teacher's Note:
1. Convert turns per cm to turns per metre before using B = mu0 n I.
2. At the end of a long solenoid the axial field is exactly half the interior value.

Answer: (b) \( 2.7\times10^{-2}\ \text{weber/m}^2 \)

Explanation:
1. \( B=\mu_0 nI=4\pi\times10^{-7}\times4250\times5=2.67\times10^{-2}\ \text{T} \).
2. This rounds to \( 2.7\times10^{-2}\ \text{Wb/m}^2 \).

Teacher's Note:
1. The length is 1 m, so n equals the number of turns.
2. Do the arithmetic as 4 pi x 4250 x 5 x 1e-7 = 2.67e-2.

Answer: (b) \( 5.6\times10^{-5}\ \text{T} \)

Explanation:
1. For circular motion, \( r=\frac{mv}{eB}\Rightarrow B=\frac{mv}{er} \).
2. \( B=\frac{9\times10^{-31}\times10^{6}}{1.6\times10^{-19}\times0.10}=5.6\times10^{-5}\ \text{T} \).

Teacher's Note:
1. The magnetic force supplies the centripetal force: evB = mv^2/r.
2. Keep mass in kg and radius in metres.

Answer: (b) Y-axis

Explanation:
1. \( \vec F=q(\vec v\times\vec B)\propto\hat k\times\hat i=\hat j \).
2. The proton is positive, so the force is along the +Y-axis.

Teacher's Note:
1. Use the right-hand rule: z cross x gives +y.
2. The force is always perpendicular to both v and B.

Answer: (d) 100 keV

Explanation:
1. From \( r=\frac{\sqrt{2mK}}{qB} \): \( K=\frac{q^2B^2r^2}{2m} \).
2. For the same q, B and r, \( K\propto\frac{1}{m} \). A deuteron has twice the mass of a proton, so the proton's energy is double: 100 keV.

Teacher's Note:
1. Deuteron and proton carry the same charge but the deuteron is twice as heavy.
2. At fixed radius and field, kinetic energy is inversely proportional to mass.

Answer: (c) \( \frac{R_1^2}{2R_2^2} \)

Explanation:
1. \( qV=\tfrac12mv^2 \) and \( r=\frac{mv}{qB} \) give \( r=\frac{1}{B}\sqrt{\frac{2mV}{q}} \), so \( m=\frac{qB^2r^2}{2V} \).
2. \( \frac{m_X}{m_Y}=\frac{q_X}{q_Y}\cdot\frac{R_1^2}{R_2^2}=\frac12\cdot\frac{R_1^2}{R_2^2}=\frac{R_1^2}{2R_2^2} \).

Teacher's Note:
1. After acceleration through V, the radius depends on sqrt(m/q).
2. Remember q_X/q_Y = 1/2 because Y has double the charge.

Answer: (d) a helix

Explanation:
1. Resolve the velocity: the component along H is unchanged (uniform motion), the component perpendicular to H gives circular motion.
2. Combined, the path is a helix.

Teacher's Note:
1. Only when the velocity is parallel to B is the path straight, and only when perpendicular is it a circle.
2. Any other angle gives a helix with the axis along B.

Answer: (d) it can go undeflected also

Explanation:
1. If the electron moves perpendicular to both fields with speed \( v=E/B \) and the directions are chosen so that the two forces oppose, the net force is zero.
2. So undeflected motion is possible (this is the velocity selector).

Teacher's Note:
1. A velocity selector uses crossed E and B fields.
2. The other options describe motions that are not guaranteed by crossed fields.

Answer: (c) zero

Explanation:
1. The magnetic force is always perpendicular to the velocity, so it does no work at any instant.
2. Hence the total work over a full circle is zero.

Teacher's Note:
1. Magnetic force never changes the speed or kinetic energy of a charge.
2. The centripetal force is perpendicular to the displacement at every point.

Answer: (d) along OY

Explanation:
1. At P (to the right of an upward current) the wire's field points into the page, i.e. along \( -\hat k \).
2. \( \vec F=Q\,\vec V\times\vec B\propto\hat i\times(-\hat k)=+\hat j \).
3. The force is along +Y, i.e. along OY.

Teacher's Note:
1. Use the right-hand grip rule for the wire's field, then the cross product for the force.
2. The force is perpendicular to the velocity, which rules out the options along or opposite to OX.

Answer: (a) \( 5\times10^{-5}\ \text{N/m} \)

Explanation:
1. \( \frac{F}{l}=\frac{\mu_0 I_1I_2}{2\pi d}=\frac{2\times10^{-7}\times5\times5}{0.1}=5\times10^{-5}\ \text{N/m} \).

Teacher's Note:
1. The force between parallel currents per metre is mu0 I1 I2 / 2 pi d.
2. mu0/2pi = 2 x 10^-7 in SI units.

Answer: (c) \( i\pi r^2 \)

Explanation:
1. Magnetic moment = current x area = \( i\,\pi r^2 \).

Teacher's Note:
1. The magnetic moment of a plane loop is IA, directed along the normal.
2. Area of a circle is pi r^2.

Answer: (c) the poles are cylindrically cut

Explanation:
1. Cylindrical pole pieces with a soft-iron core in between make the field uniform in magnitude and radial.
2. The coil's plane is then always parallel to the field and the deflecting torque is NIAB for every angle.

Teacher's Note:
1. Radial field means the torque does not depend on the deflection angle.
2. This makes the deflection proportional to the current (linear scale).

Answer: (b) \( i\propto\theta \)

Explanation:
1. Deflecting torque \( NiAB \) balances the restoring torque \( k\theta \).
2. So \( i=\frac{k}{NAB}\theta \), i.e. \( i\propto\theta \).

Teacher's Note:
1. The radial field keeps the deflecting torque independent of angle.
2. A spring gives restoring torque proportional to the twist.

Answer: (d) \( \frac{\sqrt3}{4}I\,l^2B \)

Explanation:
1. Area of an equilateral triangle: \( A=\frac{\sqrt3}{4}l^2 \).
2. The plane lies along B, so the moment is perpendicular to B and \( \tau=IAB\sin90^\circ=\frac{\sqrt3}{4}Il^2B \).

Teacher's Note:
1. When the plane is parallel to B the torque is maximum.
2. Torque is linear in B (not B squared); option (c) is dimensionally wrong.

Answer: (c) 5

Explanation:
1. At the midpoint each wire gives a field proportional to its current. Same direction: fields oppose, so \( B_1-B_2=20 \). Reversed: fields add, so \( B_1+B_2=30 \).
2. Solving, \( B_1=25\ \mu\text{T},\ B_2=5\ \mu\text{T} \).
3. \( \frac{i_1}{i_2}=\frac{B_1}{B_2}=5 \).

Teacher's Note:
1. Midway means equal distances, so field is proportional to current.
2. Same-direction currents give opposite fields in the gap between them.

Answer: (a) \( \frac{\mu_0 i}{2R}\left(\frac{2\pi-\theta}{2\pi}\right)^2 \)

Explanation:
1. If the radius of the turned circle is r, the arc length is \( (2\pi-\theta)r=2\pi R \), so \( r=\frac{2\pi R}{2\pi-\theta} \).
2. \( B=\frac{\mu_0 i}{4\pi r}(2\pi-\theta)=\frac{\mu_0 i(2\pi-\theta)^2}{8\pi^2R}=\frac{\mu_0 i}{2R}\left(\frac{2\pi-\theta}{2\pi}\right)^2 \).

Teacher's Note:
1. The radius changes because the wire length is fixed: r is not R.
2. Use B = mu0 i (angle) / 4 pi r for an arc.