Question

A conducting square loop of side L moves with a uniform speed V in a region of uniform magnetic field acting perpendicular to the plane of the loop and directed into it as shown in figure. The emf induced in the loop is:

• A. BLV
• B. BLV/2
• C. zero
• D. 2BLV

Hint:

No flux change = No induced emf.

Answer 1

The Correct Option is C

Step 1: What causes induced emf.
By Faraday's law, an emf appears only when the magnetic flux through the loop is changing: \[ \varepsilon = -\frac{d\Phi}{dt}. \]
Step 2: Write the flux through the loop.
The flux is $\Phi = B \times A$, where $A = L^2$ is the loop area and $B$ is the field, which points into the plane everywhere in the region.
Step 3: See what happens as the loop moves.
The loop stays fully inside the uniform field as it moves. Its area $L^2$ does not change and the field $B$ is the same everywhere, so $\Phi = BL^2$ stays constant.
Step 4: Apply Faraday's law.
Since $\Phi$ does not change with time, $\frac{d\Phi}{dt} = 0$, and therefore \[ \varepsilon = -\frac{d\Phi}{dt} = 0. \]
Step 5: A quick physical check.
The emf on the front edge and the back edge are equal and opposite, so they cancel out around the loop. No current flows.
Step 6: Conclusion.
The induced emf in the loop is zero. \[ \boxed{\text{zero}} \]