A square loop of wire of side 2.3 cm and electrical resistance 79 ? is near a lo

Question

A square loop of wire of side 2.3 cm and electrical resistance 79 ? is near a long straight wire that carries a current of 5.9 A in the direction indicated. The long wire and loop both lie in the plane of the page. The left side of the loop is 9.0 cm from the wire.

(b) Repeat if the loop is moving to the right at a constant speed of 46 cm/s.
induced emf:
28 nV

induced current:

magnetic force:

(c) In (b), find the electric power dissipated in the loop.
______________ W

Calculate the rate at which an external force, pulling the loop to keep its speed constant, does work.
_________________ W

Explanation / Answer

here

a = 0.023m
mu0 = 1.26e-6 H/m
i = current thru wire = 5.9A
phi = total flux thru loop
x = distance from wire to LHS of loop = 0.09 m
v = speed of loop = dx/dt = 0.46 m/s
R = loop resistance = 79 ohms

b)

Magnetic force on Rhs of loop = i1*a*B(x) = i1*a*mu0*i/(2pi*x) toward wire

by putting the values in the formula

magnetic force = 354 * 10^-12 * 0.023* 1.26*10^-6 * 5.9 / ( 2 * 3.14 * 0.09)

magnetic field = 107.09 * 10^-18 N

rhs of loop = i1*a*B(x+a) = i1*a*mu0*i/[2pi*(x+a)] away from wire
net force on loop = (mu0*i*i1*a/2pi)*[a/x(x+a)] toward wire = F
then by putting the values in the formula

F = (1.26*10^-6 * 5.9 * 354 * 10^-12 * 0.023 / 2 * 3.14 ) * [ 0.023 / 0.09 ( 0.09 + 0.023)]

F = 9.638 * 0.0288 = 0.02783 *10^-18 N

c)

Power dissipated in loop = R*(i1)^2
P = 79* (354 *10^-12 )^2

P= 9899964*10^-24

Rate of work = (F*dx)/dt = F*v

Rate = 0.02783 *10^-18 * 0.46 = 0.0128 *10^-18 W

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