Class 12 Physics Electromagnetic Induction | Varying angle between magnetic field and area vector |

**Methods of producing Induced Emf in a circuit: Varying angle between magnetic field and area vector (θ)**

By changing the orientation of coil and magnetic field i.e. θ emf can be induced in the circuit.

Consider Φ = BA cosθ if θ is changed then the value of Φ will also change.

For Example:-

__Electric generator__-In case of electric generator there are rectangular coils and they are placed between the poles of the electromagnet.- When it is rotated then the orientation of the area vector of the coil and magnetic field is changing.
- As θ is changing from 0 to 90
^{0},90^{0}to 180^{0}, 180^{0}to 270^{0}and 270^{0}to 360^{0}. - As a result there is change in flux and as a result emf is induced in the generator which in turn induces current in the generator.

** Problem:- **A 1.0m long metallic rod is rotated with an angular frequency of 400rads

__Answer:-__

Length of the rod, l = 1 m

Angular frequency, ω = 400 rad/s

Magnetic field strength, B = 0.5 T

One end of the rod has zero linear velocity, while the other end has a linear velocity of l_{(i)}.

Average linear velocity of the rod, v=(l_{ (i)} +0)/2 = (l_{ (i)}/2)

Emf developed between the centre and the ring,

e=Blv =Bl (l_{ (i)}/2) = (Bl^{2}_{(i)}/2)

=(0.5x(1)^{2}x400)/2

=100V.

Hence, the emf developed between the centre and the ring is 100 V.

** Problem:- **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

__Answer:-__

Length of the rectangular wire, l = 8 cm = 0.08 m

Width of the rectangular wire, b = 2 cm = 0.02 m

Hence, area of the rectangular loop,A = lb

= 0.08 × 0.02

= 16 × 10 ^{– 4}

Magnetic field strength, B = 0.3 T

Velocity of the loop, v = 1 cm/s = 0.01 m/s

(a) Emf developed in the loop is given as:e = Blv

= 0.3x0.08x0.01 =2.4 x10^{-4}V

Time taken to travel along the width = (Distance travelled)/ (velocity)

= (b/v) = (0.02/0.01) =2s

Hence, the induced voltage is 2.4 x10^{-4}V which lasts for 2 s.

(b) Emf developed, e = Bbv

= 0.3x0.02x0.01=0.6x10^{-4}V

Time taken to travel along the length, t=(Distance travelled)/(velocity)

=(1/v) =(0.08)/(0.01) =8s.

Hence, the induced voltage is 0.6 x10^{-4} V which lasts for 8 s.

** Problem:- **Suppose the loop in above problemis stationary but the currentfeeding the electromagnet that produces the magnetic field isgradually reduced so that the field decreases from its initial valueof 0.3 T at the rate of 0.02 Ts

__Answer:-__

Initial B = 0.3 T

(dB /dt) = 0.02Ts^{-1}

ε =-(d Φ/dt) =-A (dB /dt)

Area, A = 8x10^{-2}x2x10^{-2} =16x10^{-4}m^{2}

ε =16x10^{-4}x0.02= 32x10^{-6}V

i = (ε/R) = (32x10^{-6}/1.6) =2x10^{-5} A

Power (ε i) = 6.4x10^{-10}W

Source of this power is source that produces the changing magnetic field.

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