Two long, straight, parallel wires carry currents that are directed perpendicula

Question

Two long, straight, parallel wires carry currents that are directed perpendicular to the page as shown In the figure. Wire 1 carries a current h into the page (in the negative z direction) and passes through the x axis at x = +a. Wire 2 passes through the x axis at x = -2a and carries an unknown current h. The total magnetic field at the origin due to the current-carrying wires has the magnitude 2mu_0I_1/(2pia). The current I_2 can have either of two possible values. Find the value of I_2 with the smaller magnitude, stating It In terms of I_1 and giving its direction. 2I_1 out of page 2I_2 into page 4I_1 out of page 4I_1 into page 6I_1 out of page In the figure, the current in the long, straight wire is I_1 and the wire lies in the plane of a rectangular loop, which carries a current I_2. The loop is of length l, and width a. Its left end is a distance c from the wire. Find the magnitude and direction of the net force exerted on the loop by the magnetic field created by the wire.

Explanation / Answer

magnetic field due to a long wire = B = uo*I/(2*pi*r)

r1 = 2a

r2 = 2a

B1y = +uo*I1/(2*pi*r1) = uo*I1/(2*pi*a)

if I2 is into

B2y = -uo*I2/(2*pi*r2) = -uo*I2/(4*pi*a)

given B1 + B2 = 2*uo*I1/(2*pi*a)

uo*I1/(2*pi*a) - uo*I2/(4*pi*a) = 2*uo*I1/(2*pi*a)

2I1 - I2 = 4I1

I2 = 2I1 - 4I1 = -2I1

++++++

if I2 is out of page

B2y = +uo*I2/(2*pi*r2) = -uo*I2/(4*pi*a)

given B1 + B2 = 2*uo*I1/(2*pi*a)

uo*I1/(2*pi*a) + uo*I2/(4*pi*a) = 2*uo*I1/(2*pi*a)

2I1 + I2 = 4I1

I2 = 2I1 <<<-----answer

++++++++++++++++++

5)

Fnet = I2*l*(B1 - B2)

B1 = uo*I1/(2*pi*c)

B2 = uo*I1/(2*pi*(c+a))

Fnet = (uo*I1*I2*l/(2*pi))* (1/c - 1/(c+a))

Fnet = (uo*I1*I2*l/(2*pi))* (a/(c+a))   left

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