Deriving expression for e.m.f.

Finding induced e.m.f. between the axel and rim of wheel
(a) Flux linked with the coil at any instant of time is:

‌ϕ=NBA‌cos‌ω‌t
‌‌

dϕ

dt

=NBω(−sin‌ωt)
‌ε=−‌

dϕ

dt

‌ε=NBAsin‌ωt
‌ε=ε0sin‌ωt (Here ε0=NBAω)
(b) l=‌0.5m,v=120‌rpm=2‌rps
ω=‌2πv=4π‌rad∕ s,B=4×10−4T, δ=30∘‌‌1∕2
‌BH=4×10−4×‌

√3

2

BH‌=2√3×10−4T
ε‌=‌

1

2

Bωl2
ε‌=‌

1

2

×2√3×10−4×4π×(0.5)2
ε‌=5.4×10−4‌ volt ‌
OR
● Deriving expression for magnetic energy stored in inductor and expression for energy density
● Calculating the resultant magnetic force and torque
(a) When external source supplies current to the inductor, e.m.f. is induced in it due to self induction. So the external supply has to do work to establish current. The amount of work done is :

‌dW=|ε|I‌dt‌‌(∵ε=L‌

dI

dt

)
‌dW=LI‌dt
‌W=‌

1

2

LI2
‌‌ Energy density ‌=‌

‌ Energy ‌

‌ Volume ‌

‌U=‌

‌ 1 2 LI2

‌ Volume ‌

(b)
Force of attraction experienced by the length SP of the loop per unit length
‌f1=‌

2µ0I1I2

4πr1

‌f1=‌

2×10−7×1×0.2

10×10−2

=4×10−7Nm−1
Force is attractive.
‌f2=‌

2µ0I1I2

4πr2

‌f2=‌

2×10−7×1×0.2

15×10−2

=2.6×10−7Nm−1
Force is repulsive.
So the net force experienced by the loop is (per unit length)
f=(f1−f2)
Total force experienced by the loop is :
F=(f1−f2)l=(1.4×10−7) ×5×10−2
Net force is attractive in nature.
As the lines of action of forces coincide torque is zero.