A single turn current loop in the shape of a right angle triangle with sides 5 cm, 12 cm, 13 cm is carrying a current of 2 A. The loop is in a uniform magnetic field of magnitude 0.75 T whose direction is parallel to the current in the 13 cm side of the loop. The magnitude of the magnetic force on the 5 cm side will be N. The value of is __________.
Show answer and solution
Answer: 9
Solution
The 13 cm side is parallel to B, so it experiences zero force. For the remaining two sides (5 cm and 12 cm), we can use the fact that the net force on a closed loop in a uniform field is zero, meaning F(5cm) + F(12cm) + F(13cm) = 0. Since F(13cm) = 0, we get F(5cm) = −F(12cm), so their magnitudes are equal.
For the 5 cm side: we need the component of its length perpendicular to B. The 13 cm side (hypotenuse) is along B. The 5 cm and 12 cm sides are the legs. The angle between the 12 cm side and the 13 cm hypotenuse: sin θ = 5/13 (opposite/hypotenuse). The angle between the 5 cm side and the 13 cm hypotenuse: sin φ = 12/13.
Force on 5 cm side = I × L × sin(angle between that side and B) = 2 × 0.05 × 0.75 × (12/13) = 0.075 × (12/13) = 0.9/13 = 9/130 N.
So x = 9.
For the 5 cm side: we need the component of its length perpendicular to B. The 13 cm side (hypotenuse) is along B. The 5 cm and 12 cm sides are the legs. The angle between the 12 cm side and the 13 cm hypotenuse: sin θ = 5/13 (opposite/hypotenuse). The angle between the 5 cm side and the 13 cm hypotenuse: sin φ = 12/13.
Force on 5 cm side = I × L × sin(angle between that side and B) = 2 × 0.05 × 0.75 × (12/13) = 0.075 × (12/13) = 0.9/13 = 9/130 N.
So x = 9.
[JEE Main 2023 · 24 Jan · Shift 2]