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 900,900 to 1800, 1800 to 2700 and 2700 to 3600.
  • 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-1 about an axis normal to the rod passing through its one end. The other end of the rod is contact with a metallic circular ring. A constant and uniform magnetic field of 0.5T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring?

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) = (Bl2(i)/2)

=(0.5x(1)2x400)/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–1 in a direction normal to the longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?

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 x10-4V

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

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

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

(b) Emf developed, e = Bbv

= 0.3x0.02x0.01=0.6x10-4V

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–1. If the cut is joined and the loophas a resistance of 1.6 Ω, how much power is dissipated by theloop as heat? What is the source of this power?

Answer:-

Initial B = 0.3 T

(dB /dt) = 0.02Ts-1

ε =-(d Φ/dt) =-A (dB /dt)

Area, A = 8x10-2x2x10-2 =16x10-4m2

ε =16x10-4x0.02= 32x10-6V

i = (ε/R) = (32x10-6/1.6) =2x10-5 A

 Power (ε i) = 6.4x10-10W

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

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