1.

A circular loop of metal wire of radius r is placed perpendicular to uniform magnetic field B. Half of the loop is folded about the diameter with constant angular velocity omega. If resistance of the loop is R then current in the loop is

Answer»

zero, when `theta = 0`
`(pir^(2)Bomega)/(2R)`when `theta = 0`
`(pir^(2)Bomega)/(2R)`when `theta = (pi)/(6)`
`(pir^(2)Bomega)/(4R)`when `theta = (pi)/(6)`

Solution :Magnetic flux linked with half of the loopremains constant but due to change in direction of area vector, flux linked with other half of the loop changes. Angle rotated by half loop in time t is `theta = OMEGAT`. Magnetic flux linked with the loop can be written as follows:
`phi= B xx (pi)/(2)+ Bxx (pir^(2))/(2) xx cos omegat`
Magnitude of the emf induced can be written as follows:
`E= |(dphi)/(dt)|= (Bpir^(2))/(2) xx omega sin omegat`
`e= (B omegapir^(2))/(2) sin omegat`
Or in terms of `theta` we can write it as follows:
`e= (B omegapir^(2))/(2) sin sintheta`
Current Clowing through the coil can be written as follows:
`i= (e)/(R)`
`implies i= (pir^(2)B omega)/(2R)sintheta`
We can SEE that options (a) and (d) are correct.


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