Magnetic Forces and Materials Questions and Answers

Answer: C

The flux density inside is \(\mu_{0}ke^{-s}\hat{z}\).

With no free current, ∮ H · dl = 0. Take a rectangular loop with one side inside the cylinder parallel to the axis and the other far outside. For an infinitely long cylinder with axial magnetization, H outside is zero and H inside is parallel to the axis, so H must be zero inside as well. Then

\[ \begin{aligned} \vec{B} &= \mu_{0}(\vec{H} + \vec{M}) \\ &= \mu_{0}\vec{M} = \mu_{0}ke^{-s}\hat{z} \end{aligned} \]

Zero would ignore the magnetization entirely. μ₀ke^-R uses the value of M at the surface everywhere, and the extra factor R/s has no source in this geometry: B simply follows M point by point.

Answer: D

The magnitude of the dipole moment is \(Id^{2}\).

Add and subtract the side PS (from S to P): the loop P-Q-R-S-T-U-P is then the sum of two square loops, PQRS in the \(xy\) plane and PSTU in the tilted plane, each of area \(d^{2}\), with the shared side cancelling.

PQRS, traversed P to Q (along \(+y\)) then Q to R (along \(-x\)), has moment \(Id^{2}\hat{z}\). PSTU contains PS (along \(-x\)) and PU, which leans 30° from the \(z\)-axis towards \(+y\); its moment is perpendicular to its plane:

\[ \begin{aligned} \vec{m}_{2} &= Id^{2}(\cos 30^{\circ}\hat{y} - \sin 30^{\circ}\hat{z}) \\ \vec{m} &= Id^{2}\left(\tfrac{\sqrt{3}}{2}\hat{y} + \tfrac{1}{2}\hat{z}\right) \\ |\vec{m}| &= Id^{2} \end{aligned} \]

Adding the two squares' moments as if they were perpendicular gives \(\sqrt{2}Id^{2}\), and as if parallel \(2Id^{2}\); they are in fact 120° apart, so the resultant has the magnitude of either one.

Answer: C

(A) is false but (R) is true.

(R) is true. A ferromagnetic material has a very large positive susceptibility and a very large relative permeability, so \(X_{m} \gg 0\) and \(\mu_{r} \gg 1\) is the usual description of one.

(A) is false. A paramagnetic material has a small positive susceptibility, of the order of \(10^{-5}\) to \(10^{-3}\), so that \(\mu_{r} = 1 + X_{m}\) is only just above 1. The condition given in the Assertion, \(X_{m} \gt 0\) and \(\mu_{r} \ge 1\), has no upper limit and is met by ferromagnetic materials as well, so it does not define a paramagnetic one. A diamagnetic material, by contrast, has \(X_{m} \lt 0\) and \(\mu_{r} \lt 1\).

So option B (A true, R false) and options A and D (A true) fail on the Assertion, and the Board keyed option C. A student who reads (A) loosely, as the textbook line that paramagnets have positive susceptibility, is led to option A or D; the missing size limit is what makes the Assertion wrong.

Answer: D

The angular momentum is \(BqR^{2}\).

The magnetic force provides the centripetal force for circular motion:

\[ \begin{aligned} qvB &= \frac{mv^{2}}{R} \\ mv &= qBR \end{aligned} \]

The angular momentum about the centre of the circle is

\[ L = mvR = qBR \cdot R = BqR^{2} \]

It is not zero, because the particle moves in a circle about that centre. The options with \(q^{2}\) or with an extra m have the wrong dimensions; the mass cancels when mv is replaced by qBR.

Answer: B

The motion has 2 degrees of freedom.

The charge starts with velocity along \(\hat{i}\) and the only force, \(q\overline{E}\), acts along \(\hat{j}\). Nothing ever gives it a velocity component along \(\hat{k}\), so the motion stays in the x-y plane: it moves uniformly in x and accelerates in y, tracing a parabola.

Two independent coordinates, x and y, are needed to describe it, so the number of degrees of freedom is 2. A free particle in space would have 3, motion along a line 1, and zero would mean no motion at all.

Answer: A

The trajectory is a Straight line.

The Lorentz force on the particle is \(q(\overline{E} + \overline{v} \times \overline{B})\). With \(\overline{v} = (E_{0}/B_{0})\hat{y}\) and \(\overline{B} = B_{0}\hat{x}\):

\[ \begin{aligned} \overline{v} \times \overline{B} &= \frac{E_{0}}{B_{0}}B_{0}(\hat{y} \times \hat{x}) \\ &= -E_{0}\hat{z} \end{aligned} \]

This exactly cancels \(\overline{E} = E_{0}\hat{z}\), so the net force is zero and the particle moves undeflected in a straight line. This is the principle of the velocity selector. A particle with any other speed would follow a cycloid-type path; circles and helices need a magnetic field alone.

Answer: C

The trajectory is a cycloid.

The electric field starts the particle moving from rest, and the magnetic force \(q\bar{v} \times \bar{B}\) then bends its path. The motion is the sum of a uniform drift with velocity \(\bar{v}_{d} = \frac{\bar{E} \times \bar{B}}{B^{2}}\) and a circular motion at the cyclotron frequency. Because the particle starts at rest, the circular speed equals the drift speed, so the path is that of a point on the rim of a rolling wheel: a cycloid, which momentarily stops at each cusp.

A straight line would need no magnetic force, a circle would need no electric field, and a helix needs an initial velocity component along B.

Answer: D

The SI unit is C m\(^{2}\)/s.

The magnetic dipole moment of a current loop is \(m = IA\), current times area, so its unit is the ampere square metre. Since one ampere is one coulomb per second:

\[ \begin{aligned} \text{A m}^{2} &= \frac{\text{C}}{\text{s}}\,\text{m}^{2} \\ &= \text{C m}^{2}/\text{s} \end{aligned} \]

The coulomb metre (C m) is the unit of the electric dipole moment, a common confusion. C m\(^{2}\) and C m/s lack the correct combination of area and time.

Answer: B

The torque is \(\vec{m} \times \vec{B} + \vec{r} \times \vec{\nabla}(\vec{m} \cdot \vec{B})\).

In a non-uniform field the dipole feels two things: a torque about its own centre, \(\vec{m} \times \vec{B}\), and a net force \(\vec{F} = \vec{\nabla}(\vec{m} \cdot \vec{B})\). About the origin, that force adds a moment \(\vec{r} \times \vec{F}\). The total torque about the origin is the sum.

\(\vec{m} \times \vec{B}\) alone is the torque in a uniform field or about the dipole's own centre. \(\vec{r} \times \vec{\nabla}(\vec{m} \cdot \vec{B})\) alone omits the intrinsic torque, and \(\vec{r} \times (\vec{m} \times \vec{B})\) has no physical meaning as a torque here.

Answer: C

The induced surface current density is proportional to \(I/d\).

The given result says \(K = \frac{3}{2}(\vec{B}\times\hat{n})/\mu_{0}\), so \(K\) is proportional to the field the sphere sits in. Since \(R \ll d\), the field of the wire is essentially uniform across the sphere and equal to its value at distance \(d\):

\[ \begin{aligned} B &= \frac{\mu_{0}I}{2\pi d} \\ K &\sim \frac{3}{2}\cdot\frac{I}{2\pi d} \propto \frac{I}{d} \end{aligned} \]

The radius \(R\) does not appear in the formula for \(K\), so \(I/R\) and \(I/R^{2}\) are wrong; \(1/d^{2}\) would be the fall-off of a dipole-like field, not of a long straight wire, whose field falls as \(1/d\).