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  • A rod of length l rotates with a uniform angular velocity  about an axis passing through its middle point but

A rod of length l rotates with a uniform angular velocity \omega about an axis passing through its middle point but normal to its length in a uniform magnetic field of induction B with its direction parallel to the axis of rotation. The induced emf between the two ends of the rod is:

Option: 1

\mathrm{\frac{\mathrm{B} l^2 \omega}{2}}


Option: 2

\mathrm{zero}


Option: 3

\mathrm{\left(\frac{\mathrm{B} l^2 \omega}{8}\right)}


Option: 4

\mathrm{2 \mathrm{~B} l^2 \omega}


Answers (1)

best_answer

Length of the rod between the axis of rotation and one end of the rod \mathrm{=\frac{l}{2}}

Area swept out in one rotation\mathrm{=\pi\left(\frac{l}{2}\right)^2=\left(\frac{\pi l^2}{4}\right)}

Angular velocity\mathrm{=\omega \mathrm{rads}^{-1}}

Frequency of revolution\mathrm{=\frac{\omega}{2 \pi}}

Area swept out per second\mathrm{==\frac{\pi l^2}{4}\left(\frac{\omega}{2 \pi}\right)=\frac{l^2 \omega}{8}}

Magnetic induction\mathrm{=B}

Rate of change of magnetic flux\mathrm{=\left(\frac{\mathrm{B} l^2 \omega}{8}\right)}

Magnitude of induced emf\mathrm{=\frac{B l^2 \omega}{8}}

Magnitude of induced emf between the axis and the other end is also\mathrm{\left(\frac{\mathrm{B} l^2 \omega}{8}\right)}

These two emf’s are in opposite directions. Hence, the potential difference between the two ends of the rod is zero.
 

Posted by

shivangi.bhatnagar

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