In the figure (Figure 1) , the loop is being pulled to the right at a constant s
ID: 1445746 • Letter: I
Question
In the figure (Figure 1) , the loop is being pulled to the right at a constant speed v. A constant current I flows in the long wire, in the direction shown. By looking at the emf induced in each segment of the loop due to its motion, calculate the magnitude of the net emf epsilon induced in the loop. Express your answer in terms of a, b, I, v, meu_o, r. Find the direction (clockwise or counterclockwise) of the current induced in the loop. The current is zero. The current is clockwise. The current is counterclockwise. Explain your reasoning in part B by two ways: using Lenz's law and using the magnetic force on charges in the loop. Check your answer for the emf in part A if the loop is stationary to see if it is physically reasonable. Check your answer for the emf in part A if the loop is very thin, so a right arrow 0 to see if it is physically reasonable.Explanation / Answer
(a)
Emf induced due to Upper & Bottom part of loop will cancel each other,
Net emf will be due to Left and Right part.
e1 = B1*l*v
e1 = [uo*I/(2**a)]*b*v
e2 = B2*l*v
e2 = [uo*I/(2**(r+a)]*b*v
Net emf induced in the loop,
e = e1 - e2
e = [uo*I/(2**a)]*b*v - [uo*I/(2**(r+a)]*b*v
e = (uo*I*b*v)/(2*) [ 1/a - 1/(r+a)]
(b)
Clockwise .
(c)
As the loop is moving away from the wire, the flux in the loop is decreasing in downward direction. Now current will be induced in such sa way so that it can counter the change, Therefore Current will be induced in a clockwise direction.
(d)
For v = 0
The emf indueced , e = (uo*I*b*v)/(2*) [ 1/a - 1/(r+a)] = 0
Yes this is physically reasonable as the loop is not moving, there won't be any change in flux and hence no emf will be induced.
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