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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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Knowledge Check

  • 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 . if R = 0,01 Omega , the magnitude and direction of current flowing in the loop are

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    (a) `1A`, clockwise
    B
    (b) `1A`, anticlockwise
    C
    ( c) `10 A`, clockwise
    D
    (d) `10 A`, anticlockwise

  • A loop shown in the figure is immersed in the varying magnetic field B=B_0t , directed into the page. If the total resistance of the loop is R , then the direction and magnitude of induced current in the inner circle is

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    clockwise `(B_0(pia^2-b^2))/R`
    B
    anticlockwise `(B_0pi(a^2+b^2))/R`
    C
    clockwise `(B_0(pia^2+4b^2))/R`
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  • A rectangular loop of sides 20 cm and 10 cm carries a current of 5.0 A . A uniform magnetic field of magnitude 2 T exists parallel to the longer side of the loop. The torque acting on the loop is

    A
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    B
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    C
    `2 Nm`
    D
    `3 Nm`

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