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A current carrying loop in the form of a right angle isosceles triangle ABC is placed in a uniform magnetic field acting along AB. If the magnetic force on the arm BC is F, what is the force on the arm AC?
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- - √3→F
- -→F
- →F
- √3→F
Correct Option: B
Let a current i be flowing in the loop ABC in the direction shown in the figure. If the length of each of the sides AB and BC be x then
|→F| = ixB
where B is the magnitude of the magnetic force. The direction of →F will be in the direction perpendicular to the plane of the paper and going into it.
By Pythagorus theorem, AC = √x² + x² = √2x
∴ Magnitude of force on AC
= i √2 x B sin 45°
= i√2xB × 1/√2
= ixb = |→F|
The direction of the force on AC is perpendicular to the plane of the paper and going out of it. Hence, force on AC = - →F
