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Question
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
- longer side,
- shorter side of the loop?
For how long does the induced voltage last in each case?
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Solution 1
Magnetic flux passing through an area A placed perpendicular to the magnetic field B is Φ = BA.
Let the length of the loop be l and width b and the magnitude of its velocity be. As soon as the loop is taken out of the magnetic field perpendicular to the longer side, the area bounded by the field changes, which leads to a change in From Faraday's law, the magnitude of the induced electromotive force is
|e| = `(d phi)/(dt)`
= `d/dt (BA)`
= `B ((dA)/dt)`
Now, `(dA)/dt` = lv ....(Area enclosed by the loop per second)
Hence, |e| = Blv = 0.3 weber/m2 × (8 × 10−2 m) × (10−2 m/s)
= 2.4 × 10−4 weber/second
= 2.4 × 10−4 Volt
= 0.24 milli volts
The induced potential difference will remain as long as the flux changes. Thus, the time for the potential difference |e| to persist
= `b/v`
= `(2 xx 10^-2 m)/(10^-2 m//s)`
= 2 s
From figure (b), `(dA)/(dT) = (bv)`
|e| = Bbv
= 0.3 × (2 × 10−2) × 10−2
= 0.6 × 10−4 volt
= 0.06 millivolt
|e| the duration of the = `l/v`
= `(8 xx 10^-2 m)/(10^-2 m//s)`
= 8 s
Solution 2
Here A = 8 × 2 = 16 cm2 = 16 × 10−4 m2,
B = 0.3 T
v = 1 cm s−1
Induced emf, ε = ?
i. When velocity is normal to longer side,
l = 8 cm = 8 × 10−2 m
ε = Blv = 0.3 × 8 × 10−2 × 10−2 = 2.4 × 10−4 V
Time, `t = "distance moved"/"velocity"`
= `(2 xx 10^-2)/10^-2`
= 2 sec
ii. When velocity is normal to longer side,
l = 2 cm = 2 × 10−2 m
ε = Blv = 0.3 × 8 × 10−2 × 10−2 = 2.4 × 10−4 V
Time, `t = "distance moved"/"velocity"`
= `(8 xx 10^-2)/10^-2`
= 8 sec
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