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?
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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