A wire 30 cm long with an east-west orientation carries a current of 3.0 A eastward.There is a uniform magnetic field perpendicular to this wire.If the force on the wire is 0.18 N upward,what are the direction and magnitude of the magnetic field? A)0.20 T up B)0.20 T north C)0.20 T south D)2.0 $\times$ 10^-3 T north E)2.0 $\times$ 10^-3 T up

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Quizplus · Accepted Answer

The magnitude of the magnetic field can be calculated using the formula for the magnetic force on a current-carrying conductor: F = BIL, where F is the force, B is the magnetic field strength, I is the current, and L is the length of the wire. Rearranging for B gives B = F / (I*L). Substituting the given values (F = 0.18 N, I = 3.0 A, L = 0.30 m) yields B = 0.18 / (3.0 * 0.30) = 0.20 T. Since the force is upward and the current is eastward, by the right-hand rule, the magnetic field must be directed northward to produce an upward force.

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