Question

A conducting square frame of side 'a' and a long straight wire carrying current $I$ are located in the same plane as shown in the figure. The frame moves to the right with a constant velocity 'V'. The emf induced in the frame will be proportional to

• A. $\frac{1}{(2x+a)^2}$
• B. $\frac{1}{(2x-a)(2x+a)}$
• C. $\frac{1}{(2x-a)}$
• D. $\frac{1}{(2x-a)^2}$

Answer 1

The Correct Option is B

To solve this problem, we need to determine the induced electromotive force (emf) in a conducting square frame moving parallel to a long straight wire carrying a current. The movement of the frame will result in a change in magnetic flux through the frame, which induces an emf according to Faraday's Law of Electromagnetic Induction.

Let's go step by step to derive the proportionality of the induced emf:

1. The magnetic field $B$ produced by a long straight wire carrying current $I$ at a distance $r$ is given by Ampère's Law as: $$ B = \frac{\mu_0 I}{2\pi r} $$ where $\mu_0$ is the permeability of free space.
2. Consider a small segment $dx$ of the frame at distance $x$ from the wire. The magnetic field at this segment is: $$ B(x) = \frac{\mu_0 I}{2 \pi x} $$
3. The emf induced in this segment due to its movement is: $$ d\text{emf} = B(x)\cdot v\cdot dx $$ where $v$ is the velocity of the frame.
4. To find the total induced emf, integrate this expression from $x = 2x-a$ to $x = 2x+a$ which are the positions of the sides of the square frame: $$ \text{emf} = \int_{2x-a}^{2x+a} \frac{\mu_0 I}{2 \pi x} \cdot v \cdot dx $$
5. By integrating, we get: $$ \text{emf} = \frac{\mu_0 I v}{2 \pi} \left[ \ln(x) \right]_{2x-a}^{2x+a} = \frac{\mu_0 I v}{2 \pi} \left( \ln(2x+a) - \ln(2x-a) \right) $$
6. The properties of logarithms allow us to simplify this: $$ \text{emf} \propto \ln\left(\frac{2x+a}{2x-a}\right) $$
7. Using properties of Taylor expansion or approximation for small terms, this simplifies to: $$ \text{emf} \propto \frac{1}{(2x-a)(2x+a)} $$

Thus, the emf induced in the frame is proportional to $\frac{1}{(2x-a)(2x+a)}$, which corresponds to the given correct option.

This detailed analysis shows why the correct answer is $\frac{1}{(2x-a)(2x+a)}$.