A conducting circular loop is placed in a uni | vedclass

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

Question

(D) Given: Magnetic field $B = 0.4\,T$,rate of change of radius $\frac{dr}{dt} = 1\,mm/s = 10^{-3}\,m/s$,and radius $r = 2\,cm = 2 	imes 10^{-2}\,m$.

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(D) Given: Magnetic field $B = 0.4\,T$,rate of change of radius $\frac{dr}{dt} = 1\,mm/s = 10^{-3}\,m/s$,and radius $r = 2\,cm = 2 	imes 10^{-2}\,m$.The area of the circular loop is $A = \pi r^2$.The rate of change of area is $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$.According to Faraday's law,the magnitude of induced emf is $\varepsilon = \left| \frac{d\phi}{dt} ight| = \left| \frac{d(BA)}{dt} ight| = B \frac{dA}{dt}$.Substituting the values:$\varepsilon = 0.4 	imes (2 	imes \pi 	imes 2 	imes 10^{-2} 	imes 10^{-3})\,V$$\varepsilon = 0.4 	imes 4\pi 	imes 10^{-5}\,V$$\varepsilon = 1.6\pi 	imes 10^{-5}\,V = 16\pi 	imes 10^{-6}\,V = 16\pi\,\mu V$.Since $16\pi \approx 16 	imes 3.14 = 50.24\,\mu V$,the closest integer value is $50\,\mu V$.

Similar Questions

$A$ conducting rod of length $l$ moves with a constant velocity $\upsilon$ perpendicular to a long,straight wire carrying a current $I$,as shown in the figure. Calculate the $emf$ generated between the ends of the rod.

The direction of current induced in a wire moving in a magnetic field is found using

Find out $V_A - V_B = ?$

$A$ conductor $ABCD$ is moving with speed $V$ along the $X$-axis in a region where the magnetic field is uniform and directed into the plane of the paper. Between which points is the maximum induced emf produced?

Vedclass Test Series

Exam Paper Generator

Online Exam Module