
A thin semi-circular conducting ring of radius R is falling with its plane vertical in a horizontal magnetic field B. At the position MNQ, the speed of the ring is v and the potential difference across the ring is
A) Zero
B) \[\dfrac{1}{2}Bv\pi {R^2}\] and $M$ is at higher potential
C) \[\pi RBv\] and $Q$ is at higher potential
D) \[2RBv\] and $Q$ is at higher potential

Answer
510.3k+ views
Hint: When a semi-circular ring is made to fall in a magnetic field $B$, an induced e.m.f. is produced due to the movement of electrons. Since this e.m.f. is due to the motion of the particles, so it is also called emotional e.m.f. The potential difference occurs due to this e.m.f. and the current starts flowing from higher to lower potential.
Complete step by step solution:
Step I:
Suppose the semi-circular ring is moving in XY-plane with velocity $v$ in a downward direction. The induced e.m.f. produced due to the magnetic field is given by
\[E = Bv{l_{eq}}\]---(i)
Step II:
The current flows in a direction opposite in which the force is applied. The current is flowing from $M$ to $N$ to $Q$. The length of the semi-circular wire is \[MQ = MN + NQ\]
\[l = R + R\]
\[l = 2R\]
Step III:
The value of induced e.m.f is the same in both cases.
\[{E_{net}} = {E_{MNQ}} = {E_{MQ}}\]
\[{E_{net}} = Bvl\]
Substitute value of $l$, and solve
\[{E_{net}} = Bv2R\]
\[{E_{net}} = 2BvR\]
Step IV:
It is clear that current and force act in opposite direction. Since the current in the semi-circular ring is flowing from $M$ to $Q$ and force is in opposite direction. Therefore, $Q$ is at higher potential and $M$ is at lower potential.
$\therefore $ Potential difference across the ring is \[2BvR\] and $Q$ is at a higher potential. Option (D) is the correct answer.
Note:
When the magnetic field is changing from time to time and a conductor is placed in that field, then there is a change in the electric field also. The ring is placed in changing magnetic flux and an induced emf is produced. This produces an electric current. Also if the conductor is in motion in a magnetic field then an induced e.m.f is produced which is known as motional e.m.f. The changing magnetic field produces an electric field and vice versa.
Complete step by step solution:
Step I:
Suppose the semi-circular ring is moving in XY-plane with velocity $v$ in a downward direction. The induced e.m.f. produced due to the magnetic field is given by
\[E = Bv{l_{eq}}\]---(i)
Step II:
The current flows in a direction opposite in which the force is applied. The current is flowing from $M$ to $N$ to $Q$. The length of the semi-circular wire is \[MQ = MN + NQ\]
\[l = R + R\]
\[l = 2R\]
Step III:
The value of induced e.m.f is the same in both cases.
\[{E_{net}} = {E_{MNQ}} = {E_{MQ}}\]
\[{E_{net}} = Bvl\]
Substitute value of $l$, and solve
\[{E_{net}} = Bv2R\]
\[{E_{net}} = 2BvR\]
Step IV:
It is clear that current and force act in opposite direction. Since the current in the semi-circular ring is flowing from $M$ to $Q$ and force is in opposite direction. Therefore, $Q$ is at higher potential and $M$ is at lower potential.
$\therefore $ Potential difference across the ring is \[2BvR\] and $Q$ is at a higher potential. Option (D) is the correct answer.
Note:
When the magnetic field is changing from time to time and a conductor is placed in that field, then there is a change in the electric field also. The ring is placed in changing magnetic flux and an induced emf is produced. This produces an electric current. Also if the conductor is in motion in a magnetic field then an induced e.m.f is produced which is known as motional e.m.f. The changing magnetic field produces an electric field and vice versa.
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