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A conducting circular loop of face are 2...

A conducting circular loop of face are `2.5X10^(-3) m^2` is placed perpendicular to a magnetic field which varies as `B = (0.20 T) sin[(50pi s^(-1))t]. (a)Find the chargej flowing through any cross- section during the time t=0 to t=40 ms.(b) If the resistance of the loop is `10 Omega`, find the thermal energy developed in the loop in this period.

Text Solution

Verified by Experts

The face area of the loop is `A=2
5X10^(-5) m^2 and the magnetic field changes as B = B_0 sin omega t where B_0=0
20 T and omega =50 pi s^(-1)
The resistance of the loop is R=10 Omega
The flux through the loop at time t is `Phi =B_0 A sin omega t`
The emf induced is `verepsilon =-(dPhi)/(dt) =- B_0 Aomega cos omega t
the current is i= (varepsilon)/(R)=-(B_0 A omega)(R) cos omega t =- i_0 cos omega t
The current changes sinusoidally with the time period ` T=(2pi)/(omega)=(2pi)/(50pi s^(-1)=40ms
(a) The charge flowing through any cross section in 40ms in `Q=int_0^T idt=-i_0 int_0^T cos omega t dt`=- (i_0)/(omega)[sin omega t]_0=0`
(b)The thermal energy produced in 40 ms is `H=int_0^T i^2 Rdt=i_0^2R int_0^T cos^2 omega t dt` `=(i_0 ^2 R)/(2) int_0^T(1+cos 2 omega t)dt` `=(i_0^2R)/(2)[t+(sin 2 omega t)/(2 omega)]_0^T` `=(i_0^2 RT)/(2) =(B_0^2 A^2 omega^2)/(2R^2) R((2pi)/(omega)) = (piB_0^2A^2 omega)/(R) =(pi X(0
20T)^2 X(2
5 X10^(-3)m^2)2 X(50pi s^(-1))/(10 Omega) =1
25X10^(-5) J
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