A conducting rod of length L rotates with ang | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Consider a small element of length $dr$ at a distance $r$ from the axis of rotation.The velocity of this element is $v = r\omega$.The induced emf

Vedclass · Accepted Answer

$\frac{B L^2 \omega}{2}$

Vedclass · Answer

(B) Consider a small element of length $dr$ at a distance $r$ from the axis of rotation.The velocity of this element is $v = r\omega$.The induced emf $de$ across this small element is given by $de = B v dr = B (r\omega) dr$.To find the total induced emf $e$ between the center and one end of the rod (length $L/2$),we integrate from $r = 0$ to $r = L/2$:$e = \int_{0}^{L/2} B \omega r dr = B \omega \left[ \frac{r^2}{2} \right]_{0}^{L/2} = B \omega \frac{(L/2)^2}{2} = \frac{B \omega L^2}{8}$.Since the rod rotates about its center,there are two such segments of length $L/2$ on either side of the axis,but they are connected in parallel. Thus,the potential difference between the two ends is the same as the emf induced in one segment of length $L/2$,which is $\frac{B \omega L^2}{8}$.Wait,if the question implies the rod is rotating about one end,the answer is $\frac{B \omega L^2}{2}$. Given the options and the standard interpretation of a rod of length $L$ rotating about its center,the potential difference between the two ends is $0$ because the emfs induced in the two halves cancel out. However,if the question asks for the emf between the center and one end,it is $\frac{B \omega L^2}{8}$. Looking at the provided options,it is highly likely the question assumes rotation about one end of a rod of length $L$. For a rod of length $L$ rotating about one end,$e = \int_{0}^{L} B \omega r dr = \frac{1}{2} B \omega L^2$.

$A$ conducting rod of length $L$ rotates with angular speed $\omega$ in a uniform magnetic field of induction $B$ which is perpendicular to its motion. The induced emf developed between the two ends of the rod is

Similar Questions

$A$ metallic rod of length $1 \,m$ held along the east-west direction is allowed to fall down freely. Given the horizontal component of the Earth's magnetic field $B_H = 3 \times 10^{-5} \,T$. The emf induced in the rod at an instant $t = 2 \,s$ after it is released is (Take $g = 10 \,m/s^2$):

$A$ metal conductor of length $1\;m$ rotates vertically about one of its ends at an angular velocity of $5\;rad/s$. If the horizontal component of the Earth's magnetic field is $0.2 \times 10^{-4}\;T$,then the $e.m.f.$ developed between the two ends of the conductor is:

The figure shows a square loop of side $5 \ cm$ being moved towards the right at a constant speed of $1 \ cm/s$. The front edge enters the $20 \ cm$ wide magnetic field $(B = 0.6 \ T)$ at $t = 0$. Find the $emf$ induced in the loop at $(a) \ t = 2 \ s$,$(b) \ t = 10 \ s$,and $(c) \ t = 22 \ s$.

$A$ solid metal cube of edge length $2\, cm$ is moving in a positive $y-$ direction at a constant speed of $6\, m/s$. There is a uniform magnetic field of $0.1\, T$ in the positive $z-$ direction. The potential difference between the two faces of the cube perpendicular to the $x-$ axis is.....$mV$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module