A single-turn current loop, carrying a current of 29.0 A, is in the shape of a equilateral triangle with sides 62.0 cm. The loop is in a uniform magnetic field of magnitude 52.0 mT whose direction is parallel to the current in one side of the loop. What is the magnitude of the magnetic force on either of the other sides of the loop

The magnitude of the magnetic force on either of the other sides of the loop is 0.81 N

Explanation:

Given;

current in the loop, I = 29.0 A

side of each equilateral triangle, a = 62 cm

magnitude of magnetic field, B = 52.0mT

the magnitude of the magnetic force on either of the other sides of the loop,

the angle between each side of the equilateral triangle = 60°

Magnitude of force is given as;

F = BILSinθ

F = 52 x 10⁻³ x 29 x 0.62 x Sin60

F = 52 x 10⁻³ x 29 x 0.62 x 0.866

F = 0.81 N

Therefore, the magnitude of the magnetic force on either of the other sides of the loop is 0.81 N

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer