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A vertical straight conductor X of length 0.5m is held along the positive X-axis and situated in a uniform horizontal magnetic field of 0.1T which is pointing towards the positive Y-axis. (i) Calculate the magnitude and direction of force on X, when a current of 4A is passed through it. (ii) Through what angle must X be turned in a vertical plane so that the force on X is halved

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Answer:

i. 0.2 N ii. 30°

Explanation:

(i) Calculate the magnitude and direction of force on X, when a current of 4A is passed through it.

The magnetic force F = BILsinФ where B = magnetic field strength = 0.1 T, I = current = 4 A and L= length of conductor = 0.5 m. Since the conductor X of length 0.5m is held along the positive X-axis and situated in a uniform horizontal magnetic field of 0.1T which is pointing towards the positive Y-axis, both B and L are perpendicular to each other. So, Ф = 90°

So, F = BILsinФ

F = 0.1 T × 4 A ×0.5 m × sin90°

F = 0.1 T × 4 A ×0.5 m × 1

F = 0.2 N

(ii) Through what angle must X be turned in a vertical plane so that the force on X is halved

If F' = BILsinФ' where Ф'=the new angle, and BIL = F

F'/F = sinФ'

Since F'/F = 1/2

sinФ' = 1/2

Ф' = sin⁻¹(1/2) = 30°

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