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 (x/130) N. The value of x is

Question

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 (x/130) N. The value of x is  (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/8f22034a7592ff6c355013501d386856-.png" /> Force on 5 cm side is | vecF|= text ILB sin θ =(2)(5 × 10-2) × (3/4) × (12/13)=(9/130) N So, x=9

Force on 5cm side is ∣F∣= ILB sinθ =(2)(5×10−2)×43​×1312​=1309​N So, x=9

Force on $5 c m$ side is
$∣ F ∣ = ILB sin θ$
$= (2) (5 \times 1 0^{- 2}) \times \frac{3}{4} \times \frac{12}{13} = \frac{9}{130} N$
So, $x = 9$