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Question 6.16: (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.Class 12 - Physics - Electromagnetic Induction Electromagnetic Induction Page-232

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(a) As the magnetic field will be variable with distance from long straight wire, so the FLUX through square loop can be calculated by integration.
Let us assume a width dr of square loop at a distance r from straight wire.
B=\frac{\mu_0}{4\pi}\frac{2I}{r},or,\phi=B.adr=\frac{\mu_0}{4\pi}\frac{2I}{r}a.dr\\\\As\:\:\phi=MI\implies MI=\frac{\mu_0Ia}{2\pi}log_e(1+a/x)\\\\M=\frac{\mu_0a}{2\pi}log_e(1+a/x)

Total flux associated with square loop
\phi=\int{d\phi}=\frac{\mu_0}{4\pi}2Ia\int\limits^{a+x}_x{\frac{dr}{r}}\\\\\phi=\frac{\mu_0}{4\pi}2Ia[log_er]^{x+a}_x\\\\\phi=\frac{\mu_0Ia}{2\pi}log_e(1+a/x)

(b) the square loop is MOVING right with a CONSTANT speed v,the instantaneous flux can be TAKEN as \phi=\frac{\mu_0Ia}{2\pi}log_e(1+a/x)\\\\\text{induced}\:emf,\epsilon=-\frac{d\phi}{dt}=-\frac{d\phi}{dx}\frac{dx}{dt}=-\frac{d\phi}{dt}v\\\\\epsilon=-\frac{\mu_0Iav}{2\pi}\frac{d(log_e(1+a/x))}{dx}\\\\\implies\epsilon=-\frac{\mu_0Iav}{2\pi}\frac{1}{\displaystyle{(1+a/x)}}[-a/x^2]\\\\or,\epsilon=\frac{\mu_0}{2\pi}\frac{a^2v}{x(x+a)}I\\\\\epsilon=2\times10^{-7}\frac{[0.1]^2\times10\times50}{0.2(0.2+0.1)}=1.67\times10^{-5}V


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