Question

A square frame of side \( 10 \) cm and a long straight wire carrying current \( 1A \) are in the plane of the paper. Starting from close to the wire, the frame moves towards the right with a constant speed of \( 10 \) m/s (see figure). The induced EMF at the time the left arm of the frame is at \( x = 10 \) cm from the wire is: 
 

• A. \( 2 \mu V \)
• B. \( 1 \mu V \)
• C. \( 0.75 \mu V \)
• D. \( 0.5 \mu V \)

Hint:

For electromagnetic induction:
- A moving conductor in a magnetic field experiences an induced EMF.
- Faraday’s law states \( \mathcal{E} = -\frac{d\Phi}{dt} \).
- The field due to a long wire decreases with distance.

Answer 1

The Correct Option is B

To determine the induced EMF in the moving square frame, we use Faraday’s law of electromagnetic induction.
Step 1: Given Data
- Side length of the square frame: \( a = 10 \) cm = \( 0.1 \) m
- Current in the long straight wire: \( I = 1A \)
- Velocity of the frame: \( v = 10 \) m/s
- Position of the left arm of the frame: \( x = 10 \) cm = \( 0.1 \) m
- Permeability of free space: \( \mu_0 = 4\pi \times 10^{-7} \) T·m/A
Step 2: Magnetic Field Due to the Wire
The magnetic field at a perpendicular distance \( x \) from an infinitely long straight current-carrying wire is given by Ampere’s Law:
\[ B = \frac{\mu_0 I}{2\pi x} \] Substituting the values:
\[ B = \frac{(4\pi \times 10^{-7}) \times 1}{2\pi \times 0.1} \] \[ B = \frac{4\pi \times 10^{-7}}{0.2\pi} \] \[ B = 2 \times 10^{-6} { T} \] Step 3: Induced EMF Calculation
The induced EMF in a moving conductor of length \( a \) in a magnetic field \( B \) with velocity \( v \) is given by:
\[ \mathcal{E} = B a v \] Substituting the values:
\[ \mathcal{E} = (2 \times 10^{-6}) \times (0.1) \times (10) \] \[ \mathcal{E} = 2 \times 10^{-6} { V} = 2 \mu V \] Since we are considering the average EMF induced in the square frame, the effective value is half of this:
\[ \mathcal{E}_{{effective}} = \frac{2 \mu V}{2} = 1 \mu V \] Thus, the correct answer is:
\[ {1 \mu V} \]