Question

A uniform but time varying magnetic field is present in a circular region of radius 'R'. The magnetic field is perpendicular and into the plane of loop and the magnitude of field is increasing at a constant rate \( \alpha \). There is a straight conducting rod of length 2R placed as shown in figure. The magnitude of induced emf across the rod is

• A. \( \pi R^2 \alpha \)
• B. \( \frac{1}{2} \pi R^2 \alpha \)
• C. \( \frac{1}{2} R^2 \alpha \)
• D. \( \frac{1}{4} \pi R^2 \alpha \)

Hint:

For any straight conducting rod placed as a chord in a time-varying magnetic field, the induced EMF is simply given by \( \varepsilon = \frac{1}{2} \alpha d L \), where \( d \) is the perpendicular distance of the chord from the center and \( L \) is its length. This formula works because the parallel component of the induced electric field is uniform along the chord.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem asks for the induced electromotive force (EMF) across a straight conducting rod placed in a circular region of radius \( R \) where a time-varying magnetic field \( B \) increases at a constant rate \( \alpha = \frac{dB}{dt} \).

Step 2: Key Formula or Approach:
- Faraday's Law of Induction:
\[ \oint \vec{E} \cdot d\vec{l} = -\frac{d\Phi}{dt} \]
- For a circular region, the induced electric field \( \vec{E} \) at a distance \( r \) from the center is tangential and its magnitude is given by:
\[ E(2\pi r) = \pi r^2 \frac{dB}{dt} \implies E = \frac{1}{2} r \alpha \]
- The induced EMF \( \varepsilon \) across a straight conductor is:
\[ \varepsilon = \int \vec{E} \cdot d\vec{l} \]

Step 3: Detailed Explanation:
We analyze the scenario based on the diagram, which shows a chord of length \( L = \sqrt{2}R \) subtending a \( 90^\circ \) angle at the center (connecting two perpendicular radii):
- The rod of length \( L = \sqrt{2}R \) forms the hypotenuse of a right-angled isosceles triangle with the two perpendicular radii.
- The perpendicular distance \( d \) from the center of the circular region to this chord is:
\[ d = R \cos(45^\circ) = \frac{R}{\sqrt{2}} \]
- Let the chord be oriented parallel to the x-axis at a distance \( y = d \). For any point \( (x, d) \) on the chord, the electric field is:
\[ \vec{E} = \frac{1}{2} \alpha (-d \hat{i} + x \hat{j}) \]
- The component of the electric field along the length of the rod (parallel to the x-axis) is:
\[ E_{\parallel} = \vec{E} \cdot \hat{i} = -\frac{1}{2} \alpha d \]
- Since \( E_{\parallel} \) is constant along the rod, the magnitude of the induced EMF is:
\[ \varepsilon = |E_{\parallel}| \cdot L = \left(\frac{1}{2} \alpha d\right) L \]
- Substituting \( d = \frac{R}{\sqrt{2}} \) and \( L = \sqrt{2}R \):
\[ \varepsilon = \frac{1}{2} \alpha \left(\frac{R}{\sqrt{2}}\right) (\sqrt{2}R) = \frac{1}{2} R^2 \alpha \]

Step 4: Final Answer:
The magnitude of the induced EMF across the rod is \( \frac{1}{2} R^2 \alpha \).