A) \[EndA,3\times {{10}^{-3}}mV\]
B) End A, 3 mV
C) \[End\text{ }B,3\times {{10}^{-3}}mV\]
D) End B, 3 mV
Correct Answer: D
Solution :
Induced potential difference in the rod. \[e=Bvl\] or \[e=Hvl\] Given: \[H=3\times {{10}^{-5}}Wb/{{m}^{2}},\] \[v=50m/s,\]\[l=2m\] \[\therefore \] \[e=3\times {{10}^{-5}}\times 50\times 2\] \[=3\times {{10}^{-3}}V\] \[=3mV\] Now, by Right hand palm rule. If H is towards N and rod is moving downward, then the current will move form B to A. Hence, end B will become positive.You need to login to perform this action.
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