Class 12 Physics

CBSE and UP Board

Moving Charges and Magnetism

Chapter-4 Exercise 4.14

Additional Exercise

NCERT Solutions Class 12 Physics Chapter 4 Question 14

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Radius of coil X, r₁= 16 cm = 0.16 m

Radius of coil Y, r₂ = 10 cm = 0.1 m

Number of turns of on coil X, n₁ =20

Number of turns of on coil Y, n₂ = 25

Current in coil X, I₁= 16 A

Current in coil Y, I₂= 18 A

Magnetic field due to coil X at their centre is given by the relation,

B₁ = μ0 N₁l₁/2 r₁Where,μ0 = permeability of free space = 4 π x 10⁻⁴ T m A⁻¹

B₁ = (4 π x 10⁻⁴ x 25 x 18 )/( 2 x 0.10) = 9 π x 10⁻⁴ T (Towards West)

Hence ,net magnetic field can be obtained as ,

B = B₂ -B₁ = 9 π x 10⁻⁴ T – 4 π x 10⁻⁴ T =5 π x 10⁻⁴T

= 5 x 3.14 x 10⁻⁴ T = 1.57 x 10 ⁻³ T (Towards West)