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  • Help me solve this A thin circular-conducting ring having N turns of radius R is falling with its plane vertical in a horizontal magnetic field B. At the position MNQ, the speed of ring is v, the induced e.m.f. developed across the ring is

A thin circular-conducting ring having N turns of radius R is falling with its plane vertical in a horizontal magnetic field B. At the position MNQ, the speed of ring is v, the induced e.m.f. developed across the ring is

  • Option 1)

    Zero

  • Option 2)

    \frac{BV\pi R^{2}N}{2}    and M is at higher potential

  • Option 3)

    N\pi BRv and Q is at higher potential

  • Option 4)

    2RB vN and Q is at lower potential

 

Answers (1)

As we discussed in concept

Motional EMF -

\varepsilon = Blv
 

- wherein

B\rightarrow magnetic field

l\rightarrow length

u\rightarrow velocity of u perpendicular to uniform magnetic field.

 

 Emf induced = (B.(MQ) v) N

                      = B . (2R) v N,

Q will be at lower potential.


Option 1)

Zero

This option is incorrect.

Option 2)

\frac{BV\pi R^{2}N}{2}    and M is at higher potential

This option is incorrect.

Option 3)

N\pi BRv and Q is at higher potential

This option is incorrect.

Option 4)

2RB vN and Q is at lower potential

This option is correct.

Posted by

Vakul

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