A circular conducting coil of radius 1, m is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The magnetic flux $\phi$ through the coil is given by $\phi = \vec{B} \cdot \vec{S} = B \cdot \pi R^2$.Substituting $B = \frac{4}{\pi} 	imes 10^{

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(A) The magnetic flux $\phi$ through the coil is given by $\phi = \vec{B} \cdot \vec{S} = B \cdot \pi R^2$.Substituting $B = \frac{4}{\pi} 	imes 10^{-3} (1 - \frac{t}{100})$ and $R = 1\, m$,we get $\phi = 4 	imes 10^{-3} (1 - \frac{t}{100})\, Wb$.The induced electromotive force $(EMF)$ is $\varepsilon = -\frac{d\phi}{dt} = -\frac{d}{dt} [4 	imes 10^{-3} (1 - \frac{t}{100})] = 4 	imes 10^{-5}\, V$.The magnetic field becomes zero when $1 - \frac{t}{100} = 0$,which gives $t = 100\, s$.The energy dissipated $E$ is given by $E = \frac{\varepsilon^2}{R_{coil}} 	imes t$.Substituting the values: $E = \frac{(4 	imes 10^{-5})^2}{2 	imes 10^{-6}} 	imes 100 = \frac{16 	imes 10^{-10}}{2 	imes 10^{-6}} 	imes 100 = 8 	imes 10^{-4} 	imes 100 = 0.08\, J$.Since $1\, J = 1000\, mJ$,$E = 0.08 	imes 1000 = 80\, mJ$.

Similar Questions

$A$ conducting square frame of side $a$ and a long straight wire carrying current $I$ are located in the same plane as shown in the figure. The frame moves to the right with a constant velocity $V$. The $emf$ induced in the frame will be proportional to

$A$ coil of $N$ turns and mean cross-sectional area $A$ is rotating with uniform angular velocity $\omega$ about an axis at right angle to a uniform magnetic field $B$. The induced e.m.f. $E$ in the coil will be

$A$ rod of $10 \ cm$ length is moving perpendicular to a uniform magnetic field of intensity $5 \times 10^{-4} \ Wb/m^2$. If the acceleration of the rod is $5 \ m/s^2$, then the rate of increase of induced $emf$ is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module