JEE Physics Moving Charges and Magnetism MCQs Set 15

Answer: (a) 5

Explanation:
1. Midway, each wire gives a field proportional to its current. Same direction: \( B_1-B_2=20 \) (in \( \mu T \)). Opposite direction: \( B_1+B_2=30 \).
2. Solving, \( B_1=25 \) and \( B_2=5 \).
3. Since field is proportional to current, \( i_1/i_2=25/5=5 \).

Teacher's Note:
1. Same-direction currents oppose each other between the wires; opposite currents add there.
2. Always write the two cases as a difference and a sum of the individual fields.

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

Explanation:
1. A full circle gives \( \frac{\mu_0 i}{2R} \) at the centre, in proportion to the angle subtended by the current.
2. The arc present subtends \( 2\pi-\theta \), so the field is the fraction \( \frac{2\pi-\theta}{2\pi} \) of the full-circle value.
3. Hence \( B=\frac{\mu_0 i}{2R}\cdot\frac{2\pi-\theta}{2\pi} \).

Teacher's Note:
1. Field at the centre of an arc is \( \frac{\mu_0 i\,\phi}{4\pi R} \) for subtended angle \( \phi \).
2. Check by letting \( \theta\to0 \): you must recover the full-circle result.

Answer: (d) 25.12 T

Explanation:
1. Inside a solenoid filled with a material, \( B=\mu_0\mu_r n i \).
2. Substituting: \( B=4\pi\times10^{-7}\times4000\times1000\times5 \).
3. This gives \( B=25.12 \) T.

Teacher's Note:
1. The core multiplies the empty-solenoid field by \( \mu_r \).
2. Such a high field in reality would saturate the core, but the formula applies for this problem.

Answer: (a) at distance \( \frac{3r}{\pi} \) from the centre, with its current producing a field opposite to the loop's

Explanation:
1. Field of the loop at its centre: \( \frac{\mu_0 i}{2r} \) (perpendicular to the plane).
2. Field of the straight wire at distance d: \( \frac{\mu_0(3i)}{2\pi d} \), also perpendicular to the plane, and can be made opposite by choosing the direction of the current.
3. Setting them equal: \( \frac{3}{2\pi d}=\frac{1}{2r} \) gives \( d=\frac{3r}{\pi} \).

Teacher's Note:
1. For a wire lying in the plane of the loop, its field at the centre is perpendicular to that plane, which is why the two fields can cancel.
2. The direction of the wire current decides whether they oppose.

Answer: (c) the magnetic field at any point inside the pipe is zero

Explanation:
1. Take a circular Amperian loop inside the pipe, coaxial with it.
2. No current passes through the loop, so \( \oint\vec B\cdot d\vec l=0 \).
3. By symmetry B has the same magnitude all around the loop, so B = 0 at every point inside.

Teacher's Note:
1. The same argument gives zero field inside any hollow cylindrical current distribution.
2. Outside, the pipe behaves like a thin wire along its axis.

Answer: (d) is towards east at both A and B

Explanation:
1. Take east = +x, north = +y, up = +z. The loop lies in the y-z plane and its axis is along x.
2. At the top the current points along \( -y \). The magnetic moment direction is that of \( \vec r\times d\vec l \) with \( \vec r=R\hat z \): \( \hat z\times(-\hat y)=+\hat x \), i.e. east.
3. On the axis the field of a loop is along its moment on both sides of it, so the field is towards east at both A and B.

Teacher's Note:
1. Curl the fingers of the right hand along the current; the thumb gives the axis direction on both faces of the loop.
2. The field direction on the axis does not change when you cross the plane of the loop.

Answer: (a) 9 : 16

Explanation:
1. Inside, at \( r=R/2 \) from the axis (R/2 from the surface): \( B_1=\frac{\mu_0 I r}{2\pi R^2}=\frac{\mu_0 I}{4\pi R} \).
2. Outside, at \( r=\frac{3R}{2} \): \( B_2=\frac{\mu_0 I}{2\pi(3R/2)}=\frac{\mu_0 I}{3\pi R} \).
3. Energy density is proportional to \( B^2 \): \( \frac{u_1}{u_2}=\left(\frac{1/4}{1/3}\right)^2=\frac{9}{16} \).

Teacher's Note:
1. Take care: the distance in the formula is from the axis, not from the surface.
2. Field rises linearly inside and falls as 1/r outside.

Answer: (c) zero

Explanation:
1. The current splits equally between the two paths from one corner to the opposite corner, since they are identical.
2. Each path is the mirror image of the other with currents in opposite senses about the centre, so their fields at the centre are equal and opposite.
3. The field at the centre is zero, hence the force \( q\vec v\times\vec B \) is zero.

Teacher's Note:
1. Symmetry arguments often give zero field quickly.
2. Notice also that v is along the axis, so even a small field there would be parallel to v along the axis.

Answer: (d) \( \left(\frac{R_1}{R_2}\right)^2 \)

Explanation:
1. Energy gained: \( qV=\frac12 mv^2 \), so \( v=\sqrt{\frac{2qV}{m}} \).
2. Radius: \( R=\frac{mv}{qB}=\frac1B\sqrt{\frac{2mV}{q}} \), so \( R^2\propto m \).
3. Therefore \( \frac{m_X}{m_Y}=\left(\frac{R_1}{R_2}\right)^2 \).

Teacher's Note:
1. Same q and same V gives R proportional to the square root of m.
2. This is the principle behind the mass spectrometer.

Answer: (b) zero

Explanation:
1. The magnetic force is \( q\vec v\times\vec B \).
2. Moving along the field line, \( \vec v\parallel\vec B \), so the cross product vanishes.
3. The force is zero.

Teacher's Note:
1. The magnetic force is always perpendicular to v when it exists, so options along or opposite to v can never be right.
2. Motion along the field line is unaffected.

Answer: (a) towards the proton beam

Explanation:
1. A proton beam moving east is a current towards the east.
2. An electron beam moving west is also a current towards the east, since the charge is negative.
3. Parallel currents attract, so the electron beam is deflected towards the proton beam.

Teacher's Note:
1. Treat each beam as a conventional current.
2. Parallel currents attract and antiparallel currents repel.

Answer: (d) \( \frac{mv\sin\theta}{qB} \)

Explanation:
1. Only the component perpendicular to B, \( v\sin\theta \), produces circular motion.
2. So \( qBv\sin\theta=\frac{m(v\sin\theta)^2}{r} \).
3. Hence \( r=\frac{mv\sin\theta}{qB} \).

Teacher's Note:
1. The parallel component only drives the particle along the field, making the path a helix.
2. The pitch is \( \frac{2\pi m}{qB}v\cos\theta \).

Answer: (b) both the electric field and the magnetic field may be present

Explanation:
1. Undeflected motion means the net force is zero or along the velocity.
2. If both fields are present with \( qE=qvB \) in suitable perpendicular directions the forces cancel (a velocity selector).
3. So both fields may be present; it is not true that either must be zero.

Teacher's Note:
1. Also, a magnetic field parallel to v with E = 0 gives no deflection.
2. The correct statement is the one allowing all possibilities.

Answer: (c) zero

Explanation:
1. For a closed loop in a uniform field the net force is \( i\left(\oint d\vec l\right)\times\vec B \).
2. Since \( \oint d\vec l=0 \) for a closed loop, the net force is zero.
3. (There is only a tension in the wire, trying to expand the loop.)

Teacher's Note:
1. The torque is also zero here because B is perpendicular to the plane (the moment is parallel to B).
2. Net force on any closed current loop in a uniform field is zero.

Answer: (a) \( \sqrt2 \)

Explanation:
1. Before folding: \( \mu_1=IL^2 \).
2. After folding, the loop consists of two rectangles of area \( \frac{L^2}{2} \) in perpendicular planes, with moments \( \frac{IL^2}{2} \) each, at right angles.
3. Resultant: \( \mu_2=\frac{IL^2}{2}\sqrt2=\frac{IL^2}{\sqrt2} \), so the ratio is \( \sqrt2 \).

Teacher's Note:
1. Moments add as vectors.
2. Folding reduces the effective area projected on any one plane.

Answer: (b) towards the 40 A wire

Explanation:
1. The third wire's current is antiparallel to the 10 A current, so they repel and the third wire is pushed away from the 10 A wire, i.e. towards the 40 A wire.
2. The third wire's current is parallel to the 40 A current (the 10 A and 40 A currents are opposite), so they attract: again towards the 40 A wire.
3. Both forces point the same way, so the net force is towards the 40 A wire.

Teacher's Note:
1. Parallel currents attract; antiparallel repel.
2. Both effects add here.

Answer: (c) zero

Explanation:
1. Torque: \( \vec\tau=\vec\mu\times\vec B \) with \( \mu=iA \).
2. The field is perpendicular to the plane, so \( \vec\mu\parallel\vec B \).
3. Therefore the torque is zero (it would be \( iAB=10^{-4} \) N m only if the field were in the plane).

Teacher's Note:
1. Maximum torque occurs when B lies in the plane of the loop.
2. Do not substitute numbers before checking the angle.

Answer: (a) \( 10^{-3} \) N

Explanation:
1. Force per unit length: \( F/l=iB \).
2. Substituting: \( 10\times10^{-4}=10^{-3} \) N per metre.
3. Hence the force per metre is \( 10^{-3} \) N.

Teacher's Note:
1. The line is east-west and the horizontal field is north-south, so the angle is 90 degrees for the horizontal component.
2. Direction by Fleming's left-hand rule.

Answer: (b) \( \frac{2\pi}{3t} \)

Explanation:
1. The field of a moving charge is zero when its velocity is along the line joining it to A, i.e. when the line from the particle to A is tangent to the circle.
2. From A at distance 2R, the tangent points satisfy \( \cos\phi=\frac{R}{2R} \), so they are at \( \phi=60^\circ \) on either side of the line OA, i.e. \( 120^\circ \) apart on the near side.
3. The shortest interval between successive zero instants corresponds to \( 120^\circ=\frac{2\pi}{3} \): \( \omega t=\frac{2\pi}{3} \), so \( \omega=\frac{2\pi}{3t} \).

Teacher's Note:
1. The larger gap (the far side) is \( 240^\circ \) and is not the minimum.
2. Biot-Savart for a point charge: \( \vec B\propto\vec v\times\vec r \).

Answer: (d) \( \frac{\mu_0 I}{2\pi}\frac{(y\hat i-x\hat j)}{x^2+y^2} \)

Explanation:
1. Field magnitude: \( \frac{\mu_0 I}{2\pi r} \) with \( r^2=x^2+y^2 \), directed along \( \hat I\times\hat r \).
2. For current along \( -\hat k \): \( -\hat k\times(x\hat i+y\hat j)=-x\hat j+y\hat i \).
3. So \( \vec B=\frac{\mu_0 I}{2\pi}\frac{y\hat i-x\hat j}{x^2+y^2} \).

Teacher's Note:
1. Check the point (1, 0): the field should point along \( -\hat j \) for current along \( -\hat z \).
2. The field circulates around the wire by the right-hand rule.

Answer: (a) \( \frac{\mu_0 i}{4\pi R}(\hat i-\hat k) \)

Explanation:
1. Distance from the axis: \( r=\sqrt{R^2/4+R^2/4}=\frac{R}{\sqrt2} \), which is inside the cylinder.
2. Magnitude inside: \( B=\frac{\mu_0 i r}{2\pi R^2}=\frac{\mu_0 i}{2\sqrt2\,\pi R} \).
3. Direction \( \hat j\times\hat r \) with \( \hat r=\frac{\hat i+\hat k}{\sqrt2} \) gives \( \frac{\hat j\times\hat i+\hat j\times\hat k}{\sqrt2}=\frac{\hat i-\hat k}{\sqrt2} \). So \( \vec B=\frac{\mu_0 i}{4\pi R}(\hat i-\hat k) \).

Teacher's Note:
1. Use \( \hat j\times\hat i=-\hat k \) and \( \hat j\times\hat k=\hat i \).
2. Inside a uniform current cylinder the field grows linearly with distance from the axis.

Answer: (b) it increases in proportion to \( r^2 \) for r < R (a curve that rises ever more steeply) and then stays constant for r > R

Explanation:
1. By Ampere's law the circulation equals \( \mu_0 i_{enc} \).
2. For r < R, \( i_{enc}=i\frac{r^2}{R^2} \), so the circulation grows as \( r^2 \).
3. For r > R the enclosed current is the whole current i, so the circulation is constant at \( \mu_0 i \).

Teacher's Note:
1. Do not confuse the circulation with B itself: B grows linearly then falls as 1/r.
2. The circulation depends only on the current enclosed.

Answer: (d) \( \frac{q(b-a)B}{m} \)

Explanation:
1. Inside the field the particle moves on a circle of radius \( r=\frac{mv}{qB} \), starting tangentially in the +x direction.
2. Its maximum penetration along x equals r, so to reach x = b it needs \( r\ge b-a \).
3. The minimum speed is \( v=\frac{q(b-a)B}{m} \).

Teacher's Note:
1. The path is a quarter circle at the point of maximum x-penetration.
2. Width of the region is \( b-a \).

Answer: (a) \( \ell_0+\frac{mg-i\ell B}{2K} \)

Explanation:
1. Forces on the rod: weight mg down, magnetic force \( i\ell B \) up, and spring forces totalling \( 2Kx \) up (for extension x).
2. Equilibrium: \( 2Kx+i\ell B=mg \).
3. So \( x=\frac{mg-i\ell B}{2K} \) and the spring length is \( \ell_0+\frac{mg-i\ell B}{2K} \).

Teacher's Note:
1. Two springs share the load, hence the factor 2K.
2. If the magnetic force exceeded mg the springs would be compressed.

Answer: (c) \( 4.7\times10^{-5} \) N m

Explanation:
1. Maximum torque: \( \tau=NiAB \).
2. Area: \( A=\pi(0.01)^2=3.14\times10^{-4} \) m\( ^2 \).
3. So \( \tau=300\times10^{-2}\times3.14\times10^{-4}\times5\times10^{-2}=4.7\times10^{-5} \) N m.

Teacher's Note:
1. Radius is half the diameter.
2. Maximum torque occurs when the plane of the coil is parallel to B.