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NCERT Class XII Chapter
Electromagnetic Induction
Questions With Solutions

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Question : 17
Total: 17
A line charge λ per unit length is lodged uniformly onto the rim of a wheel of mass M and radius R. The wheel has light non-conducting spokes and is free to rotate without friction about its axis. A uniform magnetic field extends over a circular region within the rim. It is given by
→
B
 = - B0
^
k
 (r ≤ a ; a < R)
= 0 (otherwise)

What is the angular velocity of the wheel after the field is suddenly switched off?
Solution:  
According to Faraday’s law of electromagnetic induction the induced emf is ε = –
dϕ
dt
Thus a relation between electric field and rate of change of flux can be established.
ε = - ∫
→
E
.
→
dl
 = -
dϕ
dt
→
E
 exist along circumference of radius ‘a’ due to change in magnetic flux.
E ∫ dl = -
d
dt
(πa2B) , E × 2πa = - πa2
dB
dt
E = -
a
2
dB
dt
 ... (i)
Linear charge density on rim is λ. So, total charge on rim Q = λ2πa ...(ii)
Electric Force on the charge
F = QE = −πa2λ
dB
dt
 ⇒ m
dv
dt
 = - πa2 λ
dB
dt
In terms of angular velocity v = Rω
m
d
dt
 (Rω) = - πa2λ
dB
dt
mR dω = - πa2λdB
dω = -
πa2λ
mR
 dB
Integrating both sides, ω = -
πa2λB
mR
As direction of angular velocity is along axis.
→
ω
 = -
λa2π
mR
B
^
k
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