|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.
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?
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)
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?
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
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?
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.