Class 12 Physics
CBSE and UP Board
Electromagnetic Induction
Chapter-6 Exercise 6.16
Additional Exercise
NCERT Solutions for Class 12 Physics Chapter 6 Question 16
(a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Fig. 6.21. (b) Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, v = 10 m/s. Calculate the induced emf in the loop at the instant when x = 0.2 m. Take a = 0.1 m and assume that the loop has a large resistance.
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Take a small element dy in the loop at a distance y from the long straight wire [as shown in the given figure).
Magnetic flux associated with element dy, dφ = BdA
Where, dA = Area of element dy = a dy
B = Magnetic field at distance y
=μ0I/2πy
I = Current in the wire
μ0= Permeability of free space = 4π x 10⁻⁷
Therefore ,dφ = (μ0Ia/2π) x (dy/y)
φ = μ0Ia/2π ⌠ dy/y { NOTE- “⌠” is sign of integration}
v tends from x to a + .x.
Therefore ,φ = μ0Ia/2π x⌠a+x dy/y
=μ0Ia/2π [loge y]xa+x
=μ0Ia/2π [loge(a+x)/x]
For mutual inductance M. the flux is given as: φ = MI
Therefore ,MI = μ0Ia/2π [loge(a/x+1)]
M=μ0a/2π [loge(a/x+1)]
Ans (b).
Emf induced in the loop, e = B’av
= (μ0I/2π)x av
Given, I = 50 A
x = 0.2 m
a = 0.1 m
v= 10 m/s
e = (4π x 10⁻⁷ x 50 x 0.1 x 10 )/(2π x 0.2)
e = 5 x 10⁻⁵ V