Question

A metal rod of length $L$ rotates about one end at origin with a uniform angular velocity $\omega$. The magnetic field radially falls off as $B(r) = B_o e^{-\lambda r}$; $\lambda$ being a positive constant. The emf induced (neglecting the centripetal force on electrons in the rod) is :

• A. $B_o \omega \left[ \frac{1}{\lambda^2} - e^{-\lambda L} \left( \frac{1}{\lambda^2} + \frac{L}{\lambda} \right) \right]$
• B. $B_o \omega \left[ \frac{1}{\lambda^2} + e^{-\lambda L} \left( \frac{1}{\lambda^2} + \frac{L}{\lambda} \right) \right]$
• C. $B_o \omega \left[ \frac{4}{\lambda^2} - e^{-2\lambda L} \left( \frac{1}{\lambda^2} + \frac{2L}{\lambda} \right) \right]$
• D. $B_o \omega \left[ \frac{3}{\lambda^2} - e^{-3\lambda L} \left( \frac{3}{\lambda^2} + \frac{L}{\lambda} \right) \right]$

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Induced emf in a rotating rod is calculated by integrating the motional emf $d\epsilon = B(r)v dr$ along the length of the rod.
: Key Formula or Approach:
$\epsilon = \int_0^L B(r) (\omega r) dr$.
Substitute $B(r) = B_o e^{-\lambda r}$.
Step 2: Detailed Explanation:
\[ \epsilon = \int_0^L (B_o e^{-\lambda r}) (\omega r) dr = B_o \omega \int_0^L r e^{-\lambda r} dr \]
Use integration by parts: $\int u dv = uv - \int v du$.
Let $u = r \implies du = dr$.
Let $dv = e^{-\lambda r} dr \implies v = \frac{e^{-\lambda r}}{-\lambda}$.
\[ \epsilon = B_o \omega \left[ r \left(\frac{e^{-\lambda r}}{-\lambda}\right) \right]_0^L - B_o \omega \int_0^L \frac{e^{-\lambda r}}{-\lambda} dr \]
\[ \epsilon = B_o \omega \left[ -\frac{L e^{-\lambda L}}{\lambda} + 0 \right] + \frac{B_o \omega}{\lambda} \left[ \frac{e^{-\lambda r}}{-\lambda} \right]_0^L \]
\[ \epsilon = B_o \omega \left[ -\frac{L e^{-\lambda L}}{\lambda} - \frac{1}{\lambda^2} (e^{-\lambda L} - 1) \right] \]
\[ \epsilon = B_o \omega \left[ \frac{1}{\lambda^2} - \frac{e^{-\lambda L}}{\lambda^2} - \frac{L e^{-\lambda L}}{\lambda} \right] \]
\[ \epsilon = B_o \omega \left[ \frac{1}{\lambda^2} - e^{-\lambda L} \left( \frac{1}{\lambda^2} + \frac{L}{\lambda} \right) \right] \]
Step 3: Final Answer:
The induced emf matches option (A).