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. The total magnetic force on the two equal sides has a magnitude of:Group of answer choices0.30 N0 N0.51 N0.41 N

the total magnetic force on the 2 sides of the wire is 0.30 N

Option a) 0.30 N is the correct answer

Explanation:

Given the data in the question;

both the sides of the right angled triangle is the same and the magnetic is in the plane of the  triangle and is perpendicular to the hypotenuse.

hence, angle between the magnetic field and both the sides of the triangle is 45 degrees.

∅ = 45°

I = 2.0 A

L = 15 cm = 0.15 m

B = 0.7 T

Total magnetic force on the 2 sides of the wire will be;

F = 2BILsin∅

we substitute

F = 2 × 0.7 × 2.0 × 0.15 × sin(45°)

F = 0.42 × 0.70710678

F = 0.2970 ≈ 0.30 N

Therefore, the total magnetic force on the 2 sides of the wire is 0.30 N

Option a) 0.30 N is the correct answer

a. are