What is the magnitude of magnetic force per unit length on a wirecarrying a current of 8 A and making an angle of 30º with thedirection of a uniform magnetic field of 0.15 T?

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Solution

Given: The current carried by the wire is 8 A, the uniform magnetic field is 0.15 T, and the angle between the wire and the magnetic field is 30°.

Magnetic force per unit length on wire is given as,

F=BIsinθ

where, the magnetic field is B, the current carried in the wire is I and angle between the wire and the magnetic field is θ.

By substituting the given values in the above equation, we get,

F=0.15×8×sin 30 0 =1.2× 1 2 =0.6 Nm -1

Therefore, magnetic force per unit length on the wire is 0.6 Nm -1 .

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