Electro Magnetic Induction And Alternating Currents Question 518
A rod OA of length $ l $ is rotating (about end O) over a conducting ring in crossed magnetic field B with constant angular velocity ω as shown in figure
Options:
A) Current flowing through the rod is $ \frac{B\omega ,l^{2}}{R} $
B) Magnetic force acting on the rod is $ \frac{3B^{2}\omega ,l^{3}}{4R} $
C) Torque due to magnetic force acting on the rod is $ \frac{3B^{2}l^{4}}{8R\omega} $
D) Magnitude of external force that acts perpendicularly at the end of the rod to maintain the constant angular speed is $ \frac{3B^{2}\omega ,l^{3}}{5R} $ .
Show Answer
Answer:
Correct Answer: C
Solution:
-
$ I=\frac{\varepsilon }{\frac{2R}{3}}=\frac{3\varepsilon }{2R} $
$ =\frac{3}{2R}\times \frac{1}{2}B\omega l^{2} $ $ =\frac{3B\omega l^{2}}{4R} $
Magnetic force $ F=\frac{3B\omega l^{2}}{4R}\times l\times B=\frac{3B^{2}\omega l^{3}}{4R} $
$ \tau =\frac{3B^{2}\omega l^{3}}{4R}\times \frac{l}{2}=\frac{3B^{2}\omega l^{4}}{8R} $
$ \therefore $Force to be applied at the end $ =\frac{3B^{2}\omega l^{3}}{8R} $ .
BETA
