Semicircular wire MN of diameter L moves with speed v perpendicularly through a uniform magnetic field of strength B.
1. Find the magnitude of the induced emf generated between M and N.
2. State whether M or N is at a higher potential.
E = BLv
The Attempt at a Solution
1. Given answer: Induced emf = BLv
However, I was wondering if you can use that equation even if it's not a closed circuit?
In the semicircular wire, there shouldn't be magnetic flux linkage. And even if there was, there is no change in magnetic flux linkage.
2. N is at a higher potential
There is opposing force that resists the motion of the wire ( is this reason correct?), which, by Fleming's left hand rule, results in an induced current (if the circuit were to be closed) flowing from M to N. Thus, N is at higher potential as a result of the induced "current".
However, I know that current flows from higher potential to lower potential. So is it right to say that in the case of an induced current, if it flows from M to N, N is at a higher potential?
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