Question

Magnetic field at a distance \(x\) from point \(C\) as shown in the figure on rod is \(B = B_0 e^{-\lambda x}\), find induced emf between ends of metallic rod. 

• A. \( \dfrac{B_0\omega}{\lambda^2}\left(1 - e^{-\lambda L}(1 + \lambda L)\right) \)
• B. \( \dfrac{B_0\omega}{\lambda^2}\left(L - e^{-\lambda L}(1 + \lambda L)\right) \)
• C. \( \dfrac{B_0\omega}{\lambda^2}\left(L - e^{-2\lambda L}(1 + \lambda L)\right) \)
• D. \( \dfrac{B_0\omega}{\lambda^2}\left(2L - e^{-\lambda L}(1 + \lambda L)\right) \)

Hint:

For variable magnetic fields: \[ \varepsilon = \int B(x)\,\omega x\,dx \] Always integrate over the length when \(B\) depends on position.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
When a conducting rod is rotated in a magnetic field, motional EMF is induced across it.
Since the magnetic field is not uniform but varies with distance $x$ from the pivot, we must consider an infinitesimally small element $dx$ of the rod and integrate over the entire length $L$.
Step 2: Key Formula or Approach:
The induced EMF $d\varepsilon$ in an element $dx$ moving with speed $v = \omega x$ in a transverse magnetic field $B(x)$ is:
\[ d\varepsilon = B(x) v dx = B(x) (\omega x) dx \]
Total EMF $\varepsilon$ is found by integrating from $x = 0$ to $x = L$.
Step 3: Detailed Explanation:
Given $B(x) = B_0 e^{-\lambda x}$.
\[ \varepsilon = \int_0^L B_0 e^{-\lambda x} (\omega x) dx = B_0 \omega \int_0^L x e^{-\lambda x} dx \]
We solve the integral using integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = x \implies du = dx$.
Let $dv = e^{-\lambda x} dx \implies v = -\frac{e^{-\lambda x}}{\lambda}$.
\[ \int_0^L x e^{-\lambda x} dx = \left[ -x \frac{e^{-\lambda x}}{\lambda} \right]_0^L - \int_0^L \left(-\frac{e^{-\lambda x}}{\lambda}\right) dx \]
\[ = \left( -L \frac{e^{-\lambda L}}{\lambda} - 0 \right) + \frac{1}{\lambda} \int_0^L e^{-\lambda x} dx \]
\[ = -L \frac{e^{-\lambda L}}{\lambda} + \frac{1}{\lambda} \left[ \frac{e^{-\lambda x}}{-\lambda} \right]_0^L \]
\[ = -L \frac{e^{-\lambda L}}{\lambda} - \frac{1}{\lambda^2} \left( e^{-\lambda L} - 1 \right) \]
\[ = \frac{1}{\lambda^2} \left( 1 - e^{-\lambda L} \right) - \frac{L e^{-\lambda L}}{\lambda} \]
Factoring out $\frac{1}{\lambda^2}$:
\[ \int_0^L x e^{-\lambda x} dx = \frac{1}{\lambda^2} \left( 1 - e^{-\lambda L} - \lambda L e^{-\lambda L} \right) = \frac{1}{\lambda^2} \left( 1 - e^{-\lambda L}(1 + \lambda L) \right) \]
Substitute this back into the EMF expression:
\[ \varepsilon = \frac{B_0 \omega}{\lambda^2} \left( 1 - e^{-\lambda L}(1 + \lambda L) \right) \]
Step 4: Final Answer:
The induced emf is B$_0\omega$(1 – e$^{-\lambda L}$(1 + $\lambda$L))/$\lambda^2$.