A 2.9 m length of straight wire carries a current of 30 A in a uniform magnetic field of 65 mT whose direction?

I suggest you use the Lorentz force law to compute the force: F = L * I X B Here L is the length of the wire, I is the current through the wire, and B is the magnetic field. The operator "X" denotes cross product. | I X B | = | I | | B | sin?

So the magnitude of the force per unit length F = L * | I X B | / L = | I | | B | sin? We have 0.25 A and 0.0255 T. And sin (30) = 1/2.

So F = 0.25 * 0.0255 * 1/2 = 0.0032 N /m = 3.2e-3 N/m.

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