A single turn current loop in the shape of a right angle triangle with side 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 x/130 N. The value of ' x ' is (a) 4 (b) 9 (c) 12 (d) 15

Answer: (b). B is along the 13 cm side; angle θ between the 5 cm side and the hypotenuse has sin θ = 12/13. F = I L B sin θ = 2×0.05×0.75×12/13 F = 0.9/13 = 9/130 N So x = 9, answer (b).

A single turn current loop in the shape of a right angle triangle… | MHT CET 2023 (12 May, Shift 1)

Q30MHT CET 2023 · 12 MAY · SHIFT 1Medium

Force on Current-Carrying Conductors

A single turn current loop in the shape of a right angle triangle with side $5\text{ cm},\ 12\text{ cm},\ 13\text{ cm}$ is carrying a current of $2\text{ A}$ . The loop is in a uniform magnetic field of magnitude $0.75\text{ T}$ whose direction is parallel to the current in the $13\text{ cm}$ side of the loop. The magnitude of the magnetic force on the $5\text{ cm}$ side will be $\frac{x}{130}\text{ N}$ . The value of ' $x$ ' is

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Answer: (b)

$B$ is along the 13 cm side; angle $\theta$ between the 5 cm side and the hypotenuse has $\sin\theta = \frac{12}{13}$ .
$F = I L B \sin\theta = 2\times0.05\times0.75\times\frac{12}{13}$
$F = \frac{0.9}{13} = \frac{9}{130}\text{ N}$
So $x = 9$ , answer (b).

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