The effects due to the interaction of a current-carrying loop with a magnetic field have many applications, some as common as the electric motor. This problem illustrates the basic principles of this interaction.

Consider a current I that flows in a plane rectangular current loop with height a = 4.00cm and horizontal sides b = 2.00cm. The loop is placed into a uniform magnetic field B? in such a way that the sides of length a are perpendicular to B? , and there is an angle ? between the sides of length b and B?.

For parts A and B, the loop is initially positioned at ?=30? .

A) Assume that the current flowing into the loop is 0.500A . If the magnitude of the magnetic field is 0.300T , what is ?, the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field?

B) What happens to the loop when it reaches the position for which ?=90?, that is, when its horizontal sides of length b are perpendicular to B?.
- The direction of rotation changes because the net torque acting on the loop causes the loop to rotate in a clockwise direction.
- The net torque acting on the loop is zero, but the loop continues to rotate in a counterclockwise direction.
- The net torque acting on the loop is zero; therefore it stops rotating.
- The net force acting on the loop is zero, so the loop must be in equilibrium.

a) τ = 1.039*10⁻⁴N-m

b) The net torque acting on the loop is zero, but the loop continues to rotate in a counterclockwise  direction.

Explanation:

A) Given

I = 0.5 A

B = 0.3 T

a = 4 cm = 0.04 m

b = 2 cm = 0.02 m

θ = 30°

The torque τ acting on a current-carrying loop of area A due to the interaction of the current I flowing  through the loop with a magnetic field of magnitude B is given by

τ = I*B*A*Sin∅

where ∅ is the angle between the normal to the loop and the direction of the magnetic field.

The area A of a rectangular loop of wire with height 4.00 cm and horizontal sides 2.00 cm can be obtained as follows

Aloop = a*b   ⇒   Aloop = 0.04 m*0.02 m = 8*10⁻⁴m²

Recalling that θ is the angle between the  sides of length b and B and if we consider the normal to the loop, the angle ∅ between the normal  and the magnetic field is given by

∅ = 90°-θ    ⇒   ∅ = 90°-30° = 60°

Then, the torque will be

τ = (0.5 A)*(0.3 T)*(8*10⁻⁴m²)*Sin60° = 1.039*10⁻⁴N-m

b) We have to get the net torque τ about the vertical axis of the current loop due to the interaction of the current  with the magnetic field.

The angle ∅ between the normal to the loop and the magnetic field when the  horizontal sides of the loop of length b are perpendicular to B is

∅ = 0°

Then

τ = (0.5 A)*(0.3 T)*(8*10⁻⁴m²)*Sin 0° = 0 N-m

We can say that the net torque acting on the loop is zero, but the loop continues to rotate in a counterclockwise  direction.

The velocity of the second mass is -0.8*v

Explanation: This is a perfectly inelastic collision, where two objects that collide move together as if they were stuck to each other.

The final velocity in this type of collision is

Vf = (m1*v1 + m2*v2)/(m1 + m2)

Where the "m"'s are the masses of each object, and the v's are the velocities.

Here we know that m1 = m2 = m, v1 = v, Vf = 0.100*v

we want to find the value of v2

0.100*v = (m*v + m*v2)/(2m) = m(v + v2)/(2m) = (v + v2)/2

0,100*v = v/2 + (v2)/2

v(0.100 - 1/2) = (v2)/2

-v*0.400*2 = v2

v2 = -v*0.8

Then we know that the second mass is moving with a velocity of -0.8*v (the minus sign means that this object is moving in the opposite direction with respect to the other object)

One interesting thing in this type of collision is that the kinetic energy does not conserve