Question

A single turn current loop in the shape of a right angle triangle with sides 5cm, 12cm, 13cm is carrying a current of 2A. The loop is in a uniform magnetic field of magnitude 0.75T 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 \(\frac{x}{130} N\). The value of x is ________ .

Hint:

For magnetic forces on a current-carrying wire, use F = ILB sinθ. Ensure L and B are in consistent units.

Answer 1

Correct Answer: 9

The force on a straight conductor in a magnetic field is given by:

\[ F = I L B \sin \theta \]

Here:

• \(I = 2 \, \text{A},\)
• \(L = 5 \, \text{cm} = 0.05 \, \text{m},\)
• \(B = 0.75 \, \text{T},\)
• \(\sin \theta = 1\) (since \(B\) is perpendicular to the 5 cm side).

Substitute the values:

\[ F = (2)(0.05)(0.75)(1) = 0.075 \, \text{N} \]

The force is given as \(\frac{x}{130} \, \text{N}\). Equating:

\[ \frac{x}{130} = 0.075 \implies x = 0.075 \cdot 130 = 9.75 \]

Thus, the value of \(x\) is 9.

Answer 2

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

Hence, The correct answer is 9.