Question

Figure shows a circular coil of N turns and radius a, connected to a battery of emf ε through a rheostat. The rheostat has a total length L and resistance R. the resistance of the coil is r. A small circular loop of radius a' and resistance r' is placed coaxially with the coil. The centre of the loop is at a distance x from the centre of the coil. In the beginning, the sliding contact of the rheostat is at the left end and then onwards it is moved towards right at a constant speed v. Find the emf induced in the small circular loop at the instant (a) the contact begins to slide and (b) it has slid through half the length of the rheostat.
Figure

Answer 1

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Answer 2

(a) The emf induced in the small circular loop at the instant the contact begins to slide is $\frac{\mu_{0} N a^{2} \pi a^{2} \varepsilon R v}{2 L\left(a^{2}+x^{2}\right)^{3 / 2}(R+r)^{2}}$

(b) The emf induced in the small circular loop at the instant ,it has slid through half the length of the rheostat is $\frac{\mu_{0} N a^{2} \pi a^{\prime 2}}{2 L\left(a^{2}+x^{2}\right)^{3 / 2}} \frac{E R v}{\left(\frac{R}{2}+r\right)^{2}}$

Explanation:

The magnetic field at the center of coil 2 due to coil 1 is given by

$B=\frac{\mu_{0} N i a^{2}}{2\left(a^{2}+x^{2}\right)^{3 / 2}}$

The flux connected to coil 2 is indicated by

$\phi=B \cdot A^{\prime}=\frac{\mu_{0} N i a^{2}}{2\left(a^{2}+x^{2}\right)^{3 / 2}} \pi a^{\prime 2}$

Now let y be the distance from its left end to the sliding point.

$\mu=\frac{d y}{d t}$

net rheostatic resistance = R

When the distance from the left end of the sliding touch is y, the rheostat resistance (R ') is given by

$R^{\prime}=\frac{R}{L} y$

The current in the coil is the distance y function traveled by the rheostat's sliding touch. It is administered by

$i=\frac{\varepsilon}{\left(\frac{R}{L} y+r\right)}$

The induced emf magnitude can be computed as:

 $e=\frac{d \phi}{d t}=\frac{\mu_{0} N a^{2} \pi a^{\prime 2}}{2\left(a^{2}+x^{2}\right)^{3 / 2}} \frac{d i}{d t}$

$e=\frac{\mu_{0} N a^{2} \pi a^{\prime 2}}{2\left(a^{2}+x^{2}\right)^{3 / 2}} \frac{d}{d t} \frac{\varepsilon}{\left(\frac{R}{L} y+r\right)}$

$e=\frac{\mu_{0} N a^{2} \pi a^{\prime 2}}{2\left(a^{2}+x^{2}\right)^{3 / 2}} \varepsilon \frac{\left(-\frac{R}{L} v\right)}{\left(\frac{R}{L} y+r\right)^{2}}$

(a) In case of  y = L,

$e=\frac{\mu_{0} N a^{2} \pi a^{\prime 2} \varepsilon R v}{2 L\left(a^{2}+x^{2}\right)^{3 / 2}(R+r)^{2}}$

(b) In case of  y = L/2,

$\frac{R}{L} y=\frac{R}{2}$

 $e=\frac{\mu_{0} N a^{2} \pi a^{\prime 2}}{2 L\left(a^{2}+x^{2}\right)^{3 / 2}} \frac{\varepsilon R v}{\left(\frac{R}{2}+r\right)^{2}}$