Q. 54.6( 20 Votes )

# What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30° with the direction of a uniform magnetic field of 0.15 T?

Answer :

Given:

Current through the wire, I = 8A

Strength of magnetic field = 0.15T

Angle between direction of magnetic field and current, θ = 30°

Force on a current carrying conductor of length L carrying current I, due to a uniform magnetic field of strength B is given by,

F = IL × B …(1)

In the equation we assume length to be unity (L = 1) to calculate the force per unit length.

From equation (1) we have,

F = IB×sin(θ) …(2)

Now by putting the values in equation (2), we get

F = 8A × 0.15T × sin30°

⇒ F = 0.6 Nm^{-1}

Hence, the magnitude of force per unit length of the wire is 0.6 Nm^{-1}

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