Two concentric coils X and Y of radii 16 cm and 10 cm lie in the same vertical plane containing N-S direction. X has 20 turns and carries 16 A. Y has 25 turns & carries 18 A. X has current in anticlockwise direction and Y has current in clockwise direction for an observer, looking at the coils facing the west. The magnitude of net magnetic field at their common centre isa)5p × 10–4T towards westb)13p × 10–4T towards eastc)13p × 10–4T towards westd)5p × 10–4T towards eastCorrect answer is option 'A'. Can you explain this answer?

Question

Two concentric coils X and Y of radii 16 cm and 10 cm lie in the same vertical plane containing N-S direction. X has 20 turns and carries 16 A. Y has 25 turns & carries 18 A. X has current in anticlockwise direction and Y has current in clockwise direction for an observer, looking at the coils facing the west. The magnitude of net magnetic field at their common centre isa)5p &times; 10&ndash;4T towards westb)13p &times; 10&ndash;4T towards eastc)13p &times; 10&ndash;4T towards westd)5p &times; 10&ndash;4T towards eastCorrect answer is option 'A'. Can you explain this answer?

Answer

And is connected to a variable resistor of 4 Ω in series with a battery of 12 V. The magnetic field at the center of the coils is 0.2 T. Calculate:

a) The induced emf in coil Y when the resistance is 2 Ω.

b) The ratio of the magnetic flux through coil Y to the magnetic flux through coil X.

c) The mutual inductance between the two coils.

Solution:

a) The induced emf in coil Y is given by Faraday's law of electromagnetic induction:

emf = -N dΦ/dt

where N is the number of turns in the coil and Φ is the magnetic flux through the coil. Since the coils are concentric, the magnetic field at the center is the same for both coils, and the magnetic flux through coil Y is proportional to the current in coil X:

Φ_Y = k Φ_X

where k is the ratio of the areas of the coils:

k = (π r_Y^2)/(π r_X^2) = (r_Y/r_X)^2 = (10/16)^2 = 0.390625

Therefore, the induced emf in coil Y is:

emf_Y = -N_Y dΦ_Y/dt = -k N_Y dΦ_X/dt

The rate of change of the magnetic flux through coil X is equal to the product of the magnetic field and the area:

dΦ_X/dt = B π r_X^2 d/dt (I_X/N_X) = B π r_X^2 (1/N_X) dI_X/dt

Substituting the given values, we get:

dΦ_X/dt = (0.2 T) π (0.16 m)^2 (1/20) (16 A/s) = 0.0802 Wb/s

Therefore, the induced emf in coil Y when the resistance is 2 Ω is:

emf_Y = -k N_Y dΦ_X/dt = -(0.390625)(25)(0.0802 V/s) = -0.781 V

b) The magnetic flux through each coil is proportional to the product of the magnetic field and the area:

Φ_X = B π r_X^2 = (0.2 T) π (0.16 m)^2 = 0.0201 Wb

Φ_Y = k Φ_X = (0.390625)(0.0201 Wb) = 0.00785 Wb

Therefore, the ratio of the magnetic flux through coil Y to the magnetic flux through coil X is:

Φ_Y/Φ_X = 0.00785/0.0201 = 0.390

c) The mutual inductance between the two coils is given by:

M = k μ_0 N_X N_Y π (r_Y^2 - r_X^2)/(2l)

where μ_0 is the permeability of free space and l is the distance between the centers of the coils. Substituting the given values, we get:

M = (0.390625)(4π × 10^-7 H/m) (20) (25) π ((0.10 m)^2 - (0.16 m)^2)/(2(0.10 m)) = 0.009

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