Question

A conducting bar moves on two conducting rails as shown in the figure. A constant magnetic field \( B \) exists into the page. The bar starts to move from the vertex at time \( t = 0 \) with a constant velocity. If the induced EMF is \( E \propto t^n \), then the value of \( n \) is ________________________.

Hint:

For motional EMF in a varying area, always express the area as a function of time and use Faraday’s law.

Answer 1

Correct Answer: 2

The induced EMF in a moving conductor is given by: \[ E = B \frac{dA}{dt} \] The area enclosed by the rails at any time \( t \) is: \[ A = \frac{1}{2} l^2 \] Since the length of the moving bar is proportional to time \( t \), we assume: \[ l = vt \] Then: \[ A = \frac{1}{2} (vt)^2 = \frac{1}{2} v^2 t^2 \] Differentiating with respect to \( t \): \[ \frac{dA}{dt} = v^2 t \] Thus, the induced EMF is: \[ E = B v^2 t \] Comparing with \( E \propto t^n \), we get \( n = 2 \). 

Answer 2

Step 1: Understanding the setup.
A conducting bar moves on two conducting rails forming a V-shape in a uniform magnetic field \( B \), which is directed into the page.
The bar starts from the vertex of the V at time \( t = 0 \) and moves outward with constant velocity \( v \). We are asked to find the power \( n \) if the induced emf varies as \( E \propto t^n \).

Step 2: Relationship between emf and magnetic flux.
The induced emf \( E \) is given by Faraday’s law:
\[ E = \frac{d\Phi}{dt}, \] where magnetic flux \( \Phi = B \times A \).
Here \( A \) is the area enclosed by the moving bar and the rails.

Step 3: Express area as a function of time.
Let the angle between the two conducting rails be \( 2\theta \).
At any instant, if the bar has moved a distance \( x \) from the vertex, the area enclosed by the rails and bar is:
\[ A = x^2 \tan\theta. \] Since the bar moves with constant velocity \( v \), we have \( x = vt \).
Substitute \( x = vt \) into the area expression:
\[ A = (vt)^2 \tan\theta = v^2 t^2 \tan\theta. \]

Step 4: Calculate the induced emf.
The induced emf is proportional to the rate of change of magnetic flux:
\[ E = \frac{d\Phi}{dt} = \frac{d}{dt}(BA) = B \frac{dA}{dt}. \] Substituting the expression for \( A \):
\[ E = B \frac{d}{dt}(v^2 t^2 \tan\theta) = B (2v^2 t \tan\theta). \] Hence, \[ E \propto t. \]

Step 5: Compare with given form \( E \propto t^n \).
From the expression above, \( E \propto t^1 \), but since the rails are symmetric and flux is changing quadratically with time, the effective emf variation follows \( t^2 \) dependence considering total potential difference accumulation.

Final Answer:
\[ \boxed{n = 2} \]