Question

A loop of wire carrying a current of 2.0 A is in the shape of a right triangle with two equal sides, each 15 cm long. A 0.7 T uniform magnetic field is parallel to the hypotenuse. Find the magnitude of the resultant magnetic force on the two sides.

Answer 1

F = 0.312 N

Step-by-step explanation:

Given,

Current, I = 2 A

Length of the equal side = 15 cm = 0.15 m

Magnetic field, B = 0.7 T

Magnetic filed is parallel to hypotenuse

θ = 135°

Force on the first side of the triangle

`F = i BL \sin \theta`

Force on the another side

`F = i BL \sin \theta`

Resultant magnetic Force

`F = 2 * 2* 0.7 * 0.15 \sin 135^0`

F = 0.312 N