1.

Metal ring of radius R is placed perpendicular to uniform magnetic field B. Magnetic field starts changing at a rate alpha

Answer»

Electric field at any POINT on the ring is zero.
Electric field at any point on the ring is `alpha (R)//(2)`
EMF induced in the ring is zero
Emf induced in the ring is `piaR^(2)`

Solution :Magnetic FLUX linked with the ring can be written as follows:
`phi= piR^(2)B`
Magnitude of emf induced in the loop can be written as follows:
`e = dphi//dt = pi R^(2) (dB//dt)=pi R^(2) alpha`
We can see that OPTION (d) is correct.
Due to SYMMETRY electric field at any point on the ring is same. If E is electric field intensity, then we can multiply electric field, with length of the circle (`2piR`) to calculate induced emf. We can equate it to the calculated value of emf as follows:
`E xx 2piR= pi R^(2)alpha`
`E= alpha R//2`
Option (b) is correct. Hence, options (b) and (d) are correct.


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