Question

100 - mm - long metal rod is placed in a uniform magnetic field with the rod length perpendicular to the field direction ( Figure 1 ) The rod moves at 0.30 m / s . and its velocity vector makes an angle of 60 ° with the rod length Part A If the magnitude of the magnetic field is 0.90 T , what is the potential difference between the two ends of the rod ?​

Answer 1

Explanation:

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Answer 2

Answer:

The potential difference between the two ends of the rod is ​$13.5\times 10^{-3} V$

Explanation:

Given length of conducting rod, $l=100 mm=0.1m$

         velocity of the rod, $v=0.30 m/s$

         angle,  $\theta= 60^{o}$

        magnitude of the magnetic field,  $B= 0.90 T$

According to Faraday's law, motional emf will be generated, when a conductor moves in a magnetic field, given by

                   $\Delta V =Bvlcos\theta$

                          $=0.90\times 0.30\times 0.1\times cos60^{o}$              

                          $=0.027 \times \frac{1}{2}=0.0135V$

                          $=13.5\times 10^{-3} V$

Therefore, the potential difference between the two ends of the rod is ​$13.5\times 10^{-3} V$