A short-circuited coil is placed in a time-va | vedclass

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

Question

(B) The induced electromotive force $(e)$ is given by Faraday's law: $e = -N \frac{d\phi}{dt} = -NA \frac{dB}{dt}$.Since $A = \pi r_c^2$ (where $r_c$

Vedclass · Accepted Answer

The same

Vedclass · Answer

(B) The induced electromotive force $(e)$ is given by Faraday's law: $e = -N \frac{d\phi}{dt} = -NA \frac{dB}{dt}$.Since $A = \pi r_c^2$ (where $r_c$ is the coil radius),$e \propto N$.The resistance of the wire is $R = ho \frac{L}{A_w} = ho \frac{N(2\pi r_c)}{\pi r^2}$,where $r$ is the wire radius.Thus,$R \propto \frac{N}{r^2}$.The power dissipated is $P = \frac{e^2}{R} \propto \frac{N^2}{N/r^2} = N r^2$.Given $N_2 = 4N_1$ and $r_2 = r_1/2$,the new power $P_2$ is:$P_2 \propto (4N_1) 	imes (r_1/2)^2 = 4N_1 	imes (r_1^2/4) = N_1 r_1^2 = P_1$.Therefore,the electrical power dissipated remains the same.

$A$ short-circuited coil is placed in a time-varying magnetic field. Electrical power is dissipated due to the current induced in the coil. If the number of turns were to be quadrupled and the wire radius halved,the electrical power dissipated would be

Similar Questions

Consider the conducting square loop shown in the figure. If the switch is closed and after some time it is opened again,then the square loop will show:

In a coil of resistance $100 \ \Omega$,a current is induced by changing the magnetic flux through it as shown in the figure. The magnitude of change in flux through the coil is......$Wb$.

The current through a coil varies according to the following graph. Plot the graph for induced $emf$ $(e)$ versus time $(t)$ for the coil.

Near a circular loop of conducting wire as shown in the figure,an electron moves along a straight line. The direction of the induced current,if any,in the loop is

The network shown in the figure is a part of a complete circuit. If at a certain instant the current $i$ is $5 \, A$ and is decreasing at the rate of $10^3 \, A/s$,then $V_A - V_B$ is......$V$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module