The magnetic moment of a rectangular loop that carries 6 A and has dimensions are 0.04 m × 0.08 m is 12 Am2. How many turns are present?

The number of turns present in the loop are 625 turns.

Explanation:

Given;

current on the rectangular loop, I = 6A

Dimension of the rectangular loop, L x B = 0.04 m × 0.08 m

Magnetic moment of the loop, Pm = 12 Am²

Area of the rectangular loop, A =  L x B = 0.04 m × 0.08 m = 0.0032 m²

The magnetic moment of plane carrying coil/loop is given as;

Pm = NIA

Where;

N is the number of turns

A is the area of the loop

I is the current on the loop

Therefore, the number of turns present in the loop are 625 turns.

not too sure thinking all accelerations are pointed down

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft