Maximum torque for current loop

[Differentiate1]

Homework Statement

A current loops is oriented in three different positions relative to a uniform magnetic field. In position 1, the plane of the loop is perpendicular to the field lines. In position 2 and 3, the plane of the loop is parallel to the field as shown. The torque on the loop is maximum in:

Homework Equations

Τ = NiABsinΦ

The Attempt at a Solution

Th equation states sin(pi/2) is when torque will be maximum, but the answer states that figures 2 and 3 are when the torque on the loop is maximum. If the plane of the loop is parallel to the field, wouldn't that mean torque will be 0?

I need clarification on why figures 2 and 3 are the answer, and not figure 1.

 

[hasankamal007]

You can use the concept of a magnetic dipole here and then apply the formula τ=MxB where M is the dipole vector and B the magnetic field vector in which the loop is kept.
Imagine current flowing in loop of figure 1 and curl your right hand fingers in the current's direction, then your thumb(M vector) will be pointing in the direction of magnetic field(B vector) shown in the same figure. Hence B vector and M vector would be parallel (or anti-parallel if current's direction taken is opposite). Thus their cross product would be zero. τ=MBsin(0°)=0

In case you want to see the torque acting without using the concept of a magnetic dipole, you can do this:
Imagine clockwise current flowing in figure 2. The left edge will have a current upwards and B at that point is rightward. Thus F=i(l X B)
would give force on left edge to be acting into the plane of the image(away from viewer). Whereas for the right edge, it will be out of the plane of the image(towards viewer). Thus, it will be like rotating a cardboard piece with your hands. Hence, Torque will be maximum in figure 2.

 

[andrevdh]

Phi is the angle between the normal to the plane of the coil and the direction of the magnetic field.
Curl the fingers of your right hand around the coil in the direction of the current.
Your thumb then points in the direction of the normal.

 

[rude man]

2 and 3 look like the same thing from different viewpoints.
Anyway, F = i L x B gives the answer.

 

[HalfLostAndSearching]

I can provide some clarification on this question. The equation Τ = NiABsinΦ represents the magnitude of the torque on a current loop, where N is the number of turns in the loop, i is the current, A is the area of the loop, B is the magnitude of the magnetic field, and Φ is the angle between the normal to the loop and the direction of the magnetic field.

In position 1, the plane of the loop is perpendicular to the field lines, which means that Φ = 90 degrees. Plugging this into the equation, we get sin(90) = 1, so the torque is maximum. This is because the angle between the normal to the loop and the magnetic field is at its maximum, resulting in the maximum torque.

In positions 2 and 3, the plane of the loop is parallel to the field, which means that Φ = 0 degrees. Plugging this into the equation, we get sin(0) = 0, so the torque is 0. This is because there is no angle between the normal to the loop and the magnetic field, resulting in no torque.

Therefore, the answer stating that figures 2 and 3 are when the torque on the loop is maximum is incorrect. The correct answer is figure 1, when the plane of the loop is perpendicular to the field lines. I would suggest double-checking the answer and bringing this to the attention of your instructor.