Electro Magnetic Induction And Alternating Currents Question 518

Question: A rod OA of length $ l $ is rotating (about end 0) over a conducting ring in crossed magnetic field B with constant angular velocity co 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}\omega ,l^{4}}{8R} $

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} $ .

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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} $ .



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