Certainly! Here’s a **40-word explanation for each part**: --- ### Part (a) The magnetic dipole moment of the square coil is given by \(\vec{\mu} = -Ia^2\hat{j}\), pointing in the negative y-direction by the right-hand rule. The torque \(\vec{\tau}\) on the coil is zero because the magnetic field is perpendicular to the moment. --- ### Part (b) After the current is turned off, the field remains steady. The magnetic flux through the coil is calculated by integrating the magnetic field over the coil’s surface. The result, using the Biot-Savart law, gives \(\Phi_B = \frac{\mu_0 I a}{2\pi} \ln(2)\). --- ### Part (c) When the current increases linearly with time, Faraday’s law induces a current in the square coil. The induced emf is \(\epsilon = -d\Phi_B/dt\), and the induced current in the coil is \(I_{\text{ind}} = \epsilon/R\), with \(R\) being the resistance. --- If you have more questions or need further clarifications, feel free to ask!