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

Concept: For a rotating rod in a magnetic field, small induced emf in an element \(dx\) is: \[ d\varepsilon = B(x)\,\omega x\,dx \] Total emf is obtained by integrating along the length of the rod.
Step 1: Write expression for small emf. \[ d\varepsilon = B_0 e^{-\lambda x}\,\omega x\,dx \]
Step 2: Integrate over the length of the rod. \[ \varepsilon = \int_0^L B_0 \omega x e^{-\lambda x} dx \] \[ \varepsilon = B_0 \omega \int_0^L x e^{-\lambda x} dx \]
Step 3: Evaluate the integral. \[ \int x e^{-\lambda x} dx = \frac{-e^{-\lambda x}}{\lambda^2}(\lambda x + 1) \] Thus, \[ \varepsilon = B_0 \omega \left[\frac{-e^{-\lambda x}}{\lambda^2}(\lambda x + 1)\right]_0^L \] \[ \varepsilon = \frac{B_0 \omega}{\lambda^2}\left(1 - e^{-\lambda L}(1 + \lambda L)\right) \] \[ \boxed{\varepsilon = \frac{B_0 \omega}{\lambda^2}\left(1 - e^{-\lambda L}(1 + \lambda L)\right)} \]