Please show all work and formula used im really trying to learn how to do this p

Question

Please show all work and formula used im really trying to learn how to do this problem

A rod of length L = 11.2 cm that is forced to move at constant speed v = 4.82 m/s along horizontal rails as shown in the Figure. The rod, rails, and connecting strip at the right form a conducting loop. The rod has resistance 0.311 fl; the rest of the loop has negligible resistance. A current = 90.6 A through the long straight wire at distance a = 7.52 mm from the loop sets up a (nonuniform) magnetic field throughout loop. Find : the magnitude of the emf the current induced in the loop. At what rate is thermal energy generated in the rod? What is the magnitude of the force that must be applied to the rod to make it move at constant speed?

Explanation / Answer

(a)
The strength of the magnetic field due to a straight wire depends on the perpendicular distance from the wire.
B = µ0i/(2r).

Magnetic Flux is given by ,
= B·dA
= a a+L µ0i/(2r) xdr
= µ0 i*x /(2) ln(a+L/a)

Now, Emf Induced is given by -
e = d/dt
e = d/dt (µ0 i*x /(2) ln(a+L/a))
e = µ0 i * (dx/dt ) /(2) ln(a+L/a))
e = µ0*i*v / (2) * ln((a+L)/a)

Substituing Values -
e = 2 * 10^-7 * 90.6 * 4.82 * ln((7.52 + 112)/ 7.52)
e =  2 * 10^-7 * 90.6 * 4.82 * 2.77
e = 2.42* 10^-4 v

Induced emf , e = 2.42* 10^-4 v

(b)
Induced Current,
Iin = e/R
Iin =(2.42 * 10^-4)/ 0.311 A
Iin = 7.78 * 10^-4 A
Induced Current, Iin = 7.78 * 10^-4 A

(c)
The dissipation rate of thermal energy in the rod is = I^2 *R
= (7.78 * 10^-4)^2 * 0.311
= 1.88* 10^-7 Watt

(d)
To move a rod at a constant speed, we have to apply a force that opposes the magnetic force on the rod.
Therefore,  the magnitude of the net magnetic force exerted on the rod, which is equal to the applied force in magnitude.

Fb = Iin µ0i/(2) * ln(a+L/a)
Fb = 7.78 * 10^-4 * 2 * 10^-7 * 90.6 * 2.77 N
Fb = 3.90 * 10^-8 N

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