A rectangular wire loop of sides 8 cm and 2 cm with a small cut ismoving out of a region of uniform magnetic field of magnitude 0.3 Tdirected normal to the loop. What is the emf developed across thecut if the velocity of the loop is 1 cm s–1 in a direction normal to the(a) longer side, (b) shorter side of the loop? For how long does theinduced voltage last in each case?

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Solution

a)

Given: The length of the rectangular wire loop is 8 cm, width of the rectangular wire loop is 2 cm, the magnetic field is 0.3 T and the velocity of the loop is 1 cm/s.

The area of the loop is given as,

A=l×b
Where, the length is l and the width is b.

By substituting the given values in the above formula, we get

A=8× 10 −2 ×2× 10 −2 =16× 10 −4 m 2
The induced emf is given as,

e=Blv
Where, the magnetic field is B and the velocity is v.

By substituting the given values in the above formula, we get

e=0.3×8× 10 −2 ×1× 10 −2 =2.4× 10 −4 V
The time taken to travel along the width is given as,

t= b v
By substituting the given values in the above formula, we get

t= 0.02 0.01 =2 s
Thus, the induced voltage is 2.4× 10 −4 V and time taken to travel along the width is 2 s.

(b)

For shorter side of the loop, the induced emf is given as,

e=Bbv
By substituting the given values in the above formula, we get

e=0.3×0.02×0.01 =0.6× 10 −4 V
The time taken to travel along the length is given as,

t= l v
By substituting the given values in the above formula, we get

t= 0.08 0.01 =8 s
Thus, the induced voltage is 0.6× 10 −4 V and the time taken to travel along the length is 8 s.

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Exercise :

[ A rectangular wire loop of sides 8 cm and 2 cm 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 cm

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? ]

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