Calculate the induced emf in the loop if the current in both the wires is changing at the rate $d i / d t$ .

Open in App

Solution

Flux $= \int_{a}^{a + b} \frac{μ i}{2 π x} b d x - \int_{d + a}^{d + a + b} \frac{μ i}{2 π x} b d x$

$= \frac{μ i}{2 π} l n (\frac{a + b}{a}) - \frac{μ i}{2 π} l n (\frac{d + a + b}{d + a})$

emf $= - \frac{d f l u x}{d t} = - [\frac{μ}{2 π} l n (\frac{a + b}{a}) - \frac{μ}{2 π} l n (\frac{d + a + b}{d + a})] \frac{d i}{d t}$

Suggest Corrections

Similar questions

Q. Two coils X and Y are linked such that emf

E

is induced in Y when the current in X is changing at the rate

I^{'} (= d I / d t)

. If a current

I_{0}

is now made to flow through Y, the flux linked with X will be

Two coaxial circular loops are shown in figure, smaller loop (radius = r) is at distance x above the larger one (Radius = R) with x >> R. Now if x is changing at a constant rate $\frac{d x}{d t}$ = v > 0 and current is flowing in the larger loop as shown then :