Ques10
Asked by rbhatt16
19th July 2018,
9:37 AM
Answered by Expert
Answer:
induced emf = rate of change of flux
let e be induced emf and φ be the flux, then we have e = -(dφ)/dt
when induced emf is generated in the loop, there is a current in the loop, hence we write, e = | (dφ)/dt | = i×R .............(1)
where i is the current and R is the resistance of loop.
If A is the area of loop and B is magnetic field induction, then magneti flux φ = A×B.
Hence eqn.(1) is rewritten as, 
.................(2)
.................(2)since area A is constant, we rewrite eqn.(2) as 
In eqn.(3), q is the charge flowing in the loop. dB is change of field,
when Field B is switched off to zero in a time dt, we can take dB = (B-0) = B.
Wire of resistance R is drawn from volume V of given metal and its resitivity is ρ.
Let l be the length of wire and a be the cross-section area
then we have R = ρ×(l/a) , since V = l×a we write Resistance as R = ρ×( l2 / V ) ....................(4)
square Loop is made of wire of length l, hence side of loop is l/4 and its area l2/16
using all these informations, we can write eqn.(3) as , 
.................(5)
.................(5)
Answered by Expert
20th July 2018,
3:44 PM
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