Question

Consider a toroid of square cross section and a long straight wire on its axis as shown in the figure. The total number of turns in the toroid is N.

• A. The mutual inductance of the system is $\frac{\mu_0 Nb}{2\pi} \ln \left(\frac{a+b}{a}\right)$.
• B. If the straight wire is displaced parallel to itself so that it remains inside the toroid but not on the axis, mutual inductance decreases.
• C. If the straight wire is displaced parallel to itself so that it remains inside the toroid but not on the axis, mutual inductance increases.
• D. If the straight wire is displaced parallel to itself so that it remains inside the toroid but not on the axis, mutual inductance remains same.

Answer 1

Answer: (A)

The mutual inductance of the system is $\frac{\mu_0 Nb}{2\pi} \ln \left(\frac{a+b}{a}\right)$.

Answer 2

The magnetic field from the wire is $B(r) = \frac{\mu_0 I}{2\pi r}$. The flux through a strip of thickness $dr$ and height $b$ is $d\Phi = B(r) \cdot (b \cdot dr) = \frac{\mu_0 I b}{2\pi r} dr$. Integrating from $a$ to $a+b$ gives the total flux through the cross-section: $\Phi_{\text{cross-section}} = \int_{a}^{a+b} \frac{\mu_0 I b}{2\pi r} dr = \frac{\mu_0 I b}{2\pi} \ln\left(\frac{a+b}{a}\right)$. The mutual inductance is $M = \frac{N \Phi_{\text{cross-section}}}{I} = \frac{\mu_0 Nb}{2\pi} \ln\left(\frac{a+b}{a}\right)$. When the wire is displaced, the average distance from the wire to the toroid increases, decreasing the magnetic field and flux linkage, thus decreasing mutual inductance.