A thin semicircular conducting ring of the radius R is falling with its plane vertical in a horizontal magnetic induction overset arrow B . At the position MNQ, the speed of the ring is v and the potential difference developed across the ring is

Question

A thin semicircular conducting ring of the radius R is falling with its plane vertical in a horizontal magnetic induction overset arrow B . At the position MNQ, the speed of the ring is v and the potential difference developed across the ring is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-7uxrl3yo9z80mysg.jpg" /> (A) Zero (B) (B π R2/2) and M is at a higher potential (C) π BRv and Q is at a higher potential (D) 2 RBv and Q is at a higher potential

Tardigrade Admin · Accepted Answer

Induced motional emf in MNQ is equivalent to the motional emf in an imaginary wire MQ i. e.,

e_{MNQ} = e_{MQ} = B v l = B v (2 R)

[l = MQ = 2 R]

V_{Q} - V_{M} = v (B) (2 R)

Therefore, the potential difference developed across the ring is

2 RB v

with Q at a higher potential.

Q. A thin semicircular conducting ring of the radius $R$ is falling with its plane vertical in a horizontal magnetic induction $B \to$ . At the position MNQ, the speed of the ring is $v$ and the potential difference developed across the ring is

Zero

$\frac{B π R ^{2}}{2}$ and $M$ is at a higher potential

$π BR v$ and $Q$ is at a higher potential

2 $RB v$ and $Q$ is at a higher potential

Induced motional emf in MNQ is equivalent to the motional emf in an imaginary wire MQ i. e.,
$e_{MNQ} = e_{MQ} = B v l = B v (2 R)$ $[l = MQ = 2 R]$

$V_{Q} - V_{M} = v (B) (2 R)$
Therefore, the potential difference developed across the ring is $2 RB v$ with Q at a higher potential.

Q. A thin semicircular conducting ring of the radius R is falling with its plane vertical in a horizontal magnetic induction B→ . At the position MNQ, the speed of the ring is v and the potential difference developed across the ring is

Zero

2BπR2​ and M is at a higher potential

πBRv and Q is at a higher potential

2 RBv and Q is at a higher potential

Induced motional emf in MNQ is equivalent to the motional emf in an imaginary wire MQ i. e., eMNQ​=eMQ​=Bvl=Bv(2R) [l=MQ=2R] VQ​−VM​=v(B)(2R) Therefore, the potential difference developed across the ring is 2RBv with Q at a higher potential.

Q. A thin semicircular conducting ring of the radius $R$ is falling with its plane vertical in a horizontal magnetic induction $B \to$ . At the position MNQ, the speed of the ring is $v$ and the potential difference developed across the ring is

$\frac{B π R ^{2}}{2}$ and $M$ is at a higher potential

$π BR v$ and $Q$ is at a higher potential

2 $RB v$ and $Q$ is at a higher potential

Induced motional emf in MNQ is equivalent to the motional emf in an imaginary wire MQ i. e.,
$e_{MNQ} = e_{MQ} = B v l = B v (2 R)$ $[l = MQ = 2 R]$

$V_{Q} - V_{M} = v (B) (2 R)$
Therefore, the potential difference developed across the ring is $2 RB v$ with Q at a higher potential.