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Question Part Points Submissions Used
Question Part Points Submissions Used
Question Part Points Submissions Used
Question Part Points Submissions Used
Question Part Points Submissions Used Question Part Points Submissions Used The long, straight wire AB shown in the following figure carries a current of I1 = 11.0 A. The rectangular loop whose long edges are parallel to the wire carries a current of I2 = 3.80 A. Find the magnitude and direction of the net force exerted on the loop by the magnetic field of the wire.

Explanation / Answer

the magnitude of the net force exerted on the loop by the magnetic field of the wire;

= [mu0*l*(I1*I2)*(1/0.026 - 1/0.1)]/(2*pi)

= (4*pi*10^-7)*0.2*11*3.8*28.46/(2*pi)

= 4.76*10^-5 N

The direction of the net force exerted on the loop by the magnetic field of the wire is towards the wire. This is because the force exerted on the parallel wire of loop at 2.6cm is greater than that on the parallel wire of loop at 10cm. Since the force on parallel wire at 2.6 cm is towards the wire, so the nete force direction is towards the wire.

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