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A square loop of side a and resistance R...

A square loop of side a and resistance R is moved in the region of uniform magnetic field B(loop remaining completely insidefield), with a velocity v through a distance x. The work done is:

A

`(Bl^(2)vx)/R`

B

`(2B^(2)l^(2)vx)/R`

C

`(4B^(2)l^(2)vx)/R`

D

none

Text Solution

Verified by Experts

The correct Answer is:
D
`phi=BA=constant`.
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Knowledge Check

  • A square loop of side a and resistance B is moved in the region of uniform magnetic field B (loop remaining completely inside field), with a velocity v through a distance x. The work done is

    A
    `(Bt^2vx)/R`
    B
    `(B^2t^2vx)/R`
    C
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    D
    None of the above

  • A rectangular loop of resistance R, and sides l and X, is pulled out of a uniform magnetic field B with a for maintaining uniform velocity of withdrawal is __

    A
    `(Bl^(2)v)/(R)`
    B
    `(B^(2)l^(2)v)/(R)`
    C
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    D
    0

  • A uniform circular loop of radius a and resistance R palced perpendicular to a uniform magnetic field B . One half of the loop is rotated about the diameter with angular velocity omega as shown in Fig. Then, the current in the loop is

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    B
    (b) `(pia^(2)Bomega)/(2R)`, when `theta` is zero
    C
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    D
    (d) `(pia^(2)Bomega)/(2R)`, when `theta = pi//2`

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