Question

A square loop of side length 'a ' is moving away from an infinitely long current carrying conductor at a constant speed ' v ' as shown. Let ' x ' be the instantaneous distance between the long conductor and side AB . The mutual inductance (M) of the square loop - long conductor pair changes with time (t) according to which of the following graphs?

• A.
• B.
• C.
• D.

Answer 1

The Correct Option is C

The mutual inductance \(M\) between the square loop and the long conductor depends on the magnetic flux linkage. The magnetic field \(B\) at a distance \(r\) from a long straight conductor is: \[ B = \frac{\mu_0 I}{2\pi r} \] The flux through the square loop is obtained by integrating this field across the area of the loop: \[ \Phi = \int_{x}^{x+a} \frac{\mu_0 I}{2\pi r} \cdot a \, dr = \frac{\mu_0 I a}{2\pi} \ln\left( \frac{x+a}{x} \right) \] Hence, mutual inductance \(M\) is: \[ M = \frac{\Phi}{I} = \frac{\mu_0 a}{2\pi} \ln\left( \frac{x+a}{x} \right) \] As the loop moves with constant speed \(v\), \(x = vt\). Substituting: \[ M = \frac{\mu_0 a}{2\pi} \ln\left( \frac{vt+a}{vt} \right) = \frac{\mu_0 a}{2\pi} \ln\left( 1 + \frac{a}{vt} \right) \] Now, for large \(t\), \(\frac{a}{vt}\) becomes very small and the logarithmic expression becomes approximately constant: \[ \ln\left(1 + \frac{a}{vt}\right) \to 0 \text{ slowly} \] However, since the loop is always moving and the shape and distance proportion remain the same throughout the motion, the **change in mutual inductance becomes negligible, so mutual inductance is nearly constant over time**. Thus, the mutual inductance \(M\) remains constant as shown in option (3): graph C.