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Electric Charges And Fields

Question
CBSEENPH12037865

(a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Fig.
(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 = 0.2m. Take a = 0.1 m and assume that the loop has a large resistance.

Solution
Consider, a strip of width dx (of the square loop) at a distance x from the wire carrying current. 

Magnetic field due to current carrying wire at a distance x from the wire is B = μ0 I2π x 

Small amount of magnetic flux associated with the strip  = B dA = μ0 I2πxa dx 

Magnetic flux linked with the square loop
                  ϕ = μ0Ia2πx = bx = a+bdxx  = μ0 Ia2πlogexx = bx= a+b 

                 ϕ = μ0Ia2πlogea+bb    = μ0Ia2πlogeab+1 

Since, Flux,  ϕ = MI we have 

                 MI = μ0 Ia2πlogeab+1 

                  M = μ0a2πlogeab+1 
which is the required expression for the mutual inductance between a long straight wire and square loop of side a .

b) Given, 
Current carried by the straight wire, I = 50 A
Velocity, v = 10 m/s
x = 0.2 m
Thus,
Induced emf in the loop

                e = B l v    = μ0 I2πxl v    = 4π×10-7×502π×0.2×0.1×10    = 5×10-5 volt