Magnetic Effects of Current and Magnetism Questions and Answers

Answer: A

The strength is increased by increasing the flow of current.

The magnetic field of an electromagnet grows with the current through its coil, with the number of turns in the coil, and with a core of soft iron, which concentrates the field.

Decreasing the number of turns weakens it. Plastic and aluminium are not magnetic materials, so a plastic or aluminium core does not strengthen the field the way soft iron does.

Answer: C

The radius will be the same.

The radius of the particle's path at a given speed is set by the magnetic field: \(r = \frac{mv}{qB}\). The oscillator frequency does not appear in it. The frequency only determines whether the electric field reverses in step with the particle (the resonance condition \(f = \frac{qB}{2\pi m}\)); it does not bend the path.

So changing the frequency of the accelerating field by itself leaves the radius of the circular path unchanged. Doubling, halving or quartering the radius would need a change in \(v\) or \(B\).

Answer: D

The resistance required is 1900 Ω, in series.

At full scale the total resistance must pass 1 mA at 2 V:

\[ \begin{aligned} G + R &= \frac{V}{I_{g}} = \frac{2}{10^{-3}} = 2000\ \Omega \\ R &= 2000 - 100 = 1900\ \Omega \end{aligned} \]

2025 Ω adds the galvanometer's resistance instead of subtracting it, and 1200 or 1525 Ω do not satisfy the full-scale condition.

Answer: E

\(\mu_{r}\) is 0.5, and the material is diamagnetic.

\[ \mu_{r} = 1 + \chi = 1 - 0.5 = 0.5 \]

A negative susceptibility (equivalently \(\mu_{r} \lt 1\)) is the defining property of a diamagnetic material: it is weakly magnetised opposite to the applied field.

Paramagnets have small positive \(\chi\) (\(\mu_{r}\) slightly above 1) and ferromagnets large positive \(\chi\). A relative permeability of 0 or -1 is impossible for an ordinary material (0 only for a perfect diamagnet, a superconductor).

Answer: A

The angle is 30°.

The force on a current-carrying wire is \(F = BIL\sin\theta\):

\[ \begin{aligned} 12 &= 5 \times 4 \times 1.2 \times \sin\theta = 24\sin\theta \\ \sin\theta &= 0.5 \Rightarrow \theta = 30^{\circ} \end{aligned} \]

At 90° the force would be the maximum, 24 N, and at 0° there would be no force. 45° and 60° give 17 N and about 20.8 N.

Answer: C

The trajectory will be a helix.

Split the velocity into a component along the field and one perpendicular to it. The magnetic force qv × B acts only on the perpendicular component, turning it into uniform circular motion; the parallel component is unaffected, so the proton drifts steadily along the field while it circles. Together these give a helix.

A circle would need no velocity along the field, and a straight line would need the velocity entirely along the field. A parabola arises in a uniform electric field, not a magnetic one.

Answer: A

The field at the centre is \(\frac{\mu_{0} I}{2r}\).

By the Biot-Savart law, every element \(dl\) of a circular loop is at the same distance \(r\) from the centre and perpendicular to the line joining them, so all contributions add along the axis:

\[ \begin{aligned} B &= \frac{\mu_{0}}{4\pi}\frac{I}{r^{2}} \oint dl = \frac{\mu_{0}}{4\pi}\frac{I}{r^{2}}\, 2\pi r \\ &= \frac{\mu_{0} I}{2r} \end{aligned} \]

\(\frac{\mu_{0} I}{2\pi r}\) is the field of a long straight wire at distance \(r\), the usual confusion. The other forms do not come from either standard result.

Answer: D

The field is \(8.8 \times 10^{-3}\ \text{T}\).

Inside a long solenoid, \(B = \mu_{0} n I\), where \(n\) is the number of turns per unit length. Here \(n = 7000\) per metre and \(I = 1\) A:

\[ \begin{aligned} B &= 4\pi \times 10^{-7} \times 7000 \times 1 \\ &= 8.8 \times 10^{-3}\ \text{T} \end{aligned} \]

\(4.4 \times 10^{-3}\) T is half that value, which is the field at the end of a long solenoid, not the middle. \(7.0 \times 10^{-4}\) T comes from using \(10^{-7}\) for \(\mu_{0}\) and forgetting the \(4\pi\), and the other values are slips in the arithmetic.

Answer: B

The magnetic field is produced by the vector source \(I\vec{dl}\).

The Biot-Savart law gives the field of a current element:

\[ \begin{aligned} d\vec{B} \\ &= \frac{\mu_{0}}{4\pi}\cdot\frac{I\,\vec{dl} \times \hat{r}}{r^{2}} \end{aligned} \]

The source is the vector current element \(I\vec{dl}\), whose direction sets the direction of the field through the cross product.

The field is directly, not inversely, proportional to the current; it falls as \(\frac{1}{r^{2}}\), so it is neither proportional to \(r^{2}\) nor independent of distance; and it is long-range, like the Coulomb field.

Answer: A

The torque is IB ab.

The torque on a current loop is \(\tau = IAB\sin\theta\), where \(\theta\) is the angle between the field and the loop's normal. If the field lies in the plane of the loop, it is perpendicular to the normal, so \(\theta = 90^{\circ}\) and the torque is a maximum:

\[ \tau = IB(ab)\sin 90^{\circ} = IBab \]

Halving it or doubling it has no basis, and IB alone leaves out the area.