Q.36. A metal rod of length L rotates about on end at origin with a uniform angular velocity . The m

Q: What is the correct answer for JEE Main Physics Q36?

The correct answer is A. The induced emf in a rotating rod in a magnetic field is given by $\varepsilon = \int_0^L (\vec{v} \times \vec{B}) \cdot d\vec{l}$. Here, $\vec{v} = \omega r \hat{\theta}$ and $\vec{B} = B_0 e^{-\lambda r} \hat{r}$. Since the magnetic field is radial and velocity is tangential, $\vec{v}$ and $\vec{B}$ are perpendicular. The element $d\vec{l}$ is along the rod, i.e., $dr \hat{r}$. So, $\varepsilon

Q: Q.36. A metal rod of length L rotates about on end at origin with a uniform angular velocity . The magnetic field radially falls off as ; being a positive constant. The emf induced (neglecting the…

The correct answer is A. The induced emf in a rotating rod in a magnetic field is given by $\varepsilon = \int_0^L (\vec{v} \times \vec{B}) \cdot d\vec{l}$. Here, $\vec{v} = \omega r \hat{\theta}$ and $\vec{B} = B_0 e^{-\lambda r} \hat{r}$. Since the magnetic field is radial and velocity is tangential, $\vec{v}$ and $\vec{B}$ are perpendicular. The element $d\vec{l}$ is along the rod, i.e., $dr \hat{r}$. So, $\varepsilon

Q: What are the options for this JEE Main Physics question?

A. $B_0 \omega \left[ \frac{1}{\lambda^2} - e^{-\lambda L} \left( \frac{L}{\lambda} + \frac{1}{\lambda^2} \right) \right]$ · B. $B_0 \omega \left[ \frac{1}{\lambda^2} - e^{-2\lambda L} \left( \frac{2L}{\lambda} + \frac{1}{\lambda^2} \right) \right]$ · C. $B_0 \omega \left[ \frac{1}{\lambda^2} - e^{-3\lambda L} \left( \frac{3L}{\lambda} + \frac{1}{\lambda^2} \right) \right]$ · D. $B_0 \omega \left[ \frac{1}{\lambda^2} - e^{-\lambda L} \left( \frac{L}{\lambda} + \frac{1}{\lambda} \right) \right]$

Question

Q.36. A metal rod of length L rotates about on end at origin with a uniform angular velocity . The magnetic field radially falls off as ;  being a positive constant. The emf induced (neglecting the centripetal force on electrons in the rod) is:

FirstInTest · Accepted Answer

The induced emf in a rotating rod in a magnetic field is given by $\varepsilon = \int_0^L (\vec{v} 	imes \vec{B}) \cdot d\vec{l}$. Here, $\vec{v} = \omega r \hat{	heta}$ and $\vec{B} = B_0 e^{-\lambda r} \hat{r}$. Since the magnetic field is radial and velocity is tangential, $\vec{v}$ and $\vec{B}$ are perpendicular. The element $d\vec{l}$ is along the rod, i.e., $dr \hat{r}$. So, $\varepsilon = \int_0^L (\omega r B_0 e^{-\lambda r}) dr = B_0 \omega \int_0^L r e^{-\lambda r} dr$. Using integration by parts $\int u dv = uv - \int v du$, with $u=r$ and $dv=e^{-\lambda r}dr$, we get $du=dr$ and $v = -\frac{1}{\lambda} e^{-\lambda r}$.
$\int_0^L r e^{-\lambda r} dr = \left[ -\frac{r}{\lambda} e^{-\lambda r} ight]_0^L - \int_0^L -\frac{1}{\lambda} e^{-\lambda r} dr = \left( -\frac{L}{\lambda} e^{-\lambda L} - 0 ight) + \frac{1}{\lambda} \left[ -\frac{1}{\lambda} e^{-\lambda r} ight]_0^L = -\frac{L}{\lambda} e^{-\lambda L} - \frac{1}{\lambda^2} (e^{-\lambda L} - e^0) = -\frac{L}{\lambda} e^{-\lambda L} - \frac{1}{\lambda^2} e^{-\lambda L} + \frac{1}{\lambda^2}$.
So, $\varepsilon = B_0 \omega \left[ \frac{1}{\lambda^2} - \frac{L}{\lambda} e^{-\lambda L} - \frac{1}{\lambda^2} e^{-\lambda L} ight] = B_0 \omega \left[ \frac{1}{\lambda^2} (1 - e^{-\lambda L}) - \frac{L}{\lambda} e^{-\lambda L} ight] = B_0 \omega \left[ \frac{1}{\lambda^2} - e^{-\lambda L} \left( \frac{L}{\lambda} + \frac{1}{\lambda^2} ight) ight]$.
This matches option (1) if we factor out $e^{-\lambda L}$ from the second term: $B_0 \omega \left[ \frac{1}{\lambda^2} - e^{-\lambda L} \left( \frac{\lambda L + 1}{\lambda^2} ight) ight] = B_0 \omega \frac{1}{\lambda^2} \left[ 1 - e^{-\lambda L} (\lambda L + 1) ight]$.
The given option (1) is $B_0 \omega \left[ \frac{1}{\lambda^2} - e^{-\lambda L} \left( \frac{L}{\lambda} + \frac{1}{\lambda^2} ight) ight]$. This is the correct form.

Answer

$B_0 \omega \left[ \frac{1}{\lambda^2} - e^{-\lambda L} \left( \frac{L}{\lambda} + \frac{1}{\lambda^2} \right) \right]$

Answer

$B_0 \omega \left[ \frac{1}{\lambda^2} - e^{-2\lambda L} \left( \frac{2L}{\lambda} + \frac{1}{\lambda^2} \right) \right]$

Answer

$B_0 \omega \left[ \frac{1}{\lambda^2} - e^{-3\lambda L} \left( \frac{3L}{\lambda} + \frac{1}{\lambda^2} \right) \right]$

Answer

$B_0 \omega \left[ \frac{1}{\lambda^2} - e^{-\lambda L} \left( \frac{L}{\lambda} + \frac{1}{\lambda} \right) \right]$

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