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CBSE Class 12 Physics 2019 Outside Delhi Set 1 Paper

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Question : 25 of 27
Marks: +1, -0
SECTION-D
(a) Derive an expression for the induced emf developed when a coil of NN turns, and area of cross-section AA, is rotated at constant angular speed ω\omega in a uniform magnetic field BB.
(b) A wheel with 100 metallic spokes each 0.5 m0.5\ \text{m} long is rotated with a speed of 120 rev/min120\ \text{rev/min} in a plane normal to the horizontal component of the Earth's magnetic field. If the resultant magnetic field at that place is 4×10−4 T4 \times 10^{-4}\ \text{T} and the angle of dip at the place is 30∘30^{\circ}, find the emf induced between the axle and the rim of the wheel.
OR
(a) Derive the expression for the magnetic energy stored in an inductor when a current II develops in it. Hence, obtain the expression for the magnetic energy density.
(b) A square loop of sides 5 cm5\ \text{cm} carrying a current of 0.2 A0.2\ \text{A} in the clockwise direction is placed at a distance of 10 cm10\ \text{cm} from an infinitely long wire carrying a current of 1 A1\ \text{A} as shown. Calculate (i) the resultant magnetic force, and (ii) the torque, if any, acting on the loop.
Solution:  
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=NBAcos⁡ωt\phi = N B A \cos \omega t
dϕdt=NBω(−sin⁡ωt)\frac{d\phi}{dt} = N B \omega (-\sin \omega t)
ϵ=−dϕdt\epsilon = -\frac{d\phi}{dt}
ϵ=NBAsin⁡ωt\epsilon = N B A \sin \omega t
ϵ=ϵ0sin⁡ωt\epsilon = \epsilon_0 \sin \omega t (Here ϵ0=NBAω)\epsilon_0 = N B A \omega)
(b) l=0.5 m, v=120 rpm=2 rpsl = 0.5\ \text{m},\ v = 120\ \text{rpm} = 2\ \text{rps}
ω=2πv=4π rad/s, B=4×10−4 T,\omega = 2\pi v = 4\pi\ \text{rad/s},\ B = 4 \times 10^{-4}\ \text{T}, δ=30∘ 12\delta = 30^{\circ}\ \frac{1}{2}
BH=4×10−4×32B_H = 4 \times 10^{-4} \times \frac{\sqrt{3}}{2}
BH=23×10−4 TB_H = 2\sqrt{3} \times 10^{-4}\ \text{T}
ϵ=12Bωl2\epsilon = \frac{1}{2} B \omega l^2
ϵ=12×23×10−4×4π×(0.5)2\epsilon = \frac{1}{2} \times 2\sqrt{3} \times 10^{-4} \times 4\pi \times (0.5)^2
ϵ=5.4×10−4 volt\epsilon = 5.4 \times 10^{-4}\ \text{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=LdIdt)dW = |\epsilon| I dt\ \ \left(\because \epsilon = L \frac{dI}{dt}\right)
dW=LIdtdW = L I dt
W=12LI2W = \frac{1}{2} L I^2
Energy density=EnergyVolume\text{Energy density} = \frac{\text{Energy}}{\text{Volume}}
U=12LI2VolumeU = \frac{\frac{1}{2} L I^2}{\text{Volume}}
(b)
Force of attraction experienced by the length SP of the loop per unit length
f1=2μ0I1I24πr1f_1 = \frac{2 \mu_0 I_1 I_2}{4 \pi r_1}
f1=2×10−7×1×0.210×10−2f_1 = \frac{2 \times 10^{-7} \times 1 \times 0.2}{10 \times 10^{-2}} =4×10−7 Nm−1= 4 \times 10^{-7}\ \text{Nm}^{-1}
Force is attractive.
f2=2μ0I1I24πr2f_2 = \frac{2 \mu_0 I_1 I_2}{4 \pi r_2}
f2=2×10−7×1×0.215×10−2f_2 = \frac{2 \times 10^{-7} \times 1 \times 0.2}{15 \times 10^{-2}} =2.6×10−7 Nm−1= 2.6 \times 10^{-7}\ \text{Nm}^{-1}
Force is repulsive.
So the net force experienced by the loop is (per unit length)
f=(f1−f2)f = (f_1 - f_2)
Total force experienced by the loop is :
F=(f1−f2)l=(1.4×10−7)F = (f_1 - f_2) l = (1.4 \times 10^{-7}) ×5×10−2\times 5 \times 10^{-2}
Net force is attractive in nature.
As the lines of action of forces coincide torque is zero.
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