Name: Magnetic Flux (Part-2)
Uploaded: 2019-07-19
Duration: 6 min 14 s
Description: This video enables students to understand magnetic flux for closed objects and curved surfaces placed in the non-uniform magnetic field. Click to learn more.

Question

A rectangular wire loop of sides

8 c m

and

2 c m

with a small cut is moving out of a region of uniform magnetic field of magnitude

0.3 T

directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is

1 c m s^{- 1}

in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?

Accepted Answer

(a)

When the loop is moving normal to the long side,
emf induced in the parts $C B$ , $D E$ is zero as their lengths are parallel to the velocity, and also emf induced in $E B$ is zero as it is out of the magnetic field.
Hence emf induced in the loop (across $A F$ ) = emf induced across $C D = ε = v \times B \times l$
$\Rightarrow ε = 1 \times 10^{- 2} \times 0.3 \times 8 \times 10^{- 2}$
$\Rightarrow ε = 2.4 \times 10^{- 4} V$
After 2 seconds, the side $C D$ comes out of the magnetic field.
Hence the emf lasts for 2 seconds
(b)

When the loop is moving normal to the shorter side,
emf induced in the parts $E B$ , C $D$ is zero as their lengths are parallel to the velocity, and also emf induced in $B C$ is zero as it is out of the magnetic field.
Hence emf induced in the loop (across $A F$ ) = emf induced across $E D = ε = v \times B \times l$
$\Rightarrow ε = 1 \times 10^{- 2} \times 0.3 \times 2 \times 10^{- 2}$
$\Rightarrow ε = 0.6 \times 10^{- 4} V$
After 8 seconds, the side $D E$ comes out of the magnetic field.
Hence the emf lasts for 8 seconds

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