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A flexible circular loop 20cm in diamete...

A flexible circular loop `20cm` in diameter lies in a magneic field with magnitude `1.0 T`, direction lies into the plane of the page as shown in Fig. 3.187. The loop is pulled at the points indicated by the arrws, forming a loop of zero area in `0.314s`.

The average induced emf in the circuit is

A

(a) `0.2 V`

B

(b) `0.1sV`

C

( c) `1 V`

D

(d) `10 V`

Text Solution

Verified by Experts

The correct Answer is:
B
(b)
`epsilon_(av) = -(DeltaPhi_(B))/(Deltat) = -B(DeltaA)/(Deltat) = -B((-pir)^(2))/(Deltat)`
Since the flux through the loop is decreasing, the induced current must produce a field that goes into the page. Therefore, the current flows in clockwise direction
`I_(av) = (epsilon_(av) )/(R ) = (0.1)/(0.01) = 10 A`
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