The case that has the highest torque on the loop

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The discussion centers on calculating the torque on a current loop in a magnetic field using the formulas T = IAB sin θ and T = μ × B. The correct angle θ to use is between the loop's normal vector and the magnetic field, which is 90° for options A, B, and C, leading to a torque value of T = 2 I a b B. The initial reasoning presented was flawed as it only considered one side of the rectangle and did not account for the magnetic moment of the entire loop. Ultimately, the torque calculation should reflect the whole loop's magnetic moment rather than individual sides.
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Homework Statement


viber_image.jpg


Homework Equations


T = IAB sin θ ; θ is the angle between B and I.
T = μ × B

The Attempt at a Solution


A) T = IAB sin θ
T = √2 a b I B (θ = 45°)
B) T = 0 , because θ = 0
C) T = 0 , because θ =180°
D) T = μ × B
The direction of μ = -k , because the fingers points to the direction of current and the thump points to the direction of T.
T = (-k) × (-k)
=0
E) T = (k)×(k)
=0
So , the answer is 'A' .
Is it correct ?
 

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Unfortunately, no. :)

In your analysis of A, B, and C, you're only accounting for one side of each rectangle--the one labeled with an "I" in the diagram. In fact, in the formula ##| \mathbf{\tau} | = IAB \sin(\theta)##, the angle ##\theta## is meant to be taken between the current loop's normal vector and the magnetic field, *not* between the current and ##\mathbf{B}##. This is actually a consequence of the second formula you gave, ##\mathbf{\tau} = \mathbf{\mu} \times \mathbf{B}##, since the magnetic moment of a current loop is ##IA \mathbf{n}## (where ##\mathbf{n}## is the positively-oriented unit norm to the loop).

So actually, ##\theta = 90^{\circ}## in each of choices A, B, and C. Given this, what should the answer be?

EDIT: Your answer was correct, but your reasoning was not. Apologies for the lack of clarity.
 
Last edited:
VKint said:
So actually, ##\theta = 90^{\circ}## in each of choices A, B, and C.
And in 'D' and 'E' , ##\theta = 0^{\circ}## ?
 
VKint said:
In your analysis of A, B, and C, you're only accounting for one side of each rectangle--the one labeled with an "I" in the diagram.
Should I multiply by 4 ?
 
In D and E, ##\theta## is either ##0## or ##180^{\circ}##; in either case the torque vanishes.

Multiplying by 4 won't solve the problem. The issue is that the magnetic field makes a different angle with each of the sides of the loop. The correct way to account for this is by using the magnetic moment of the loop *as a whole* instead of trying to add up the torques on each side.
 
VKint said:
The correct way to account for this is by using the magnetic moment of the loop *as a whole*
T = μ × B
= I A B sin θ
The area is 2a*b and the angle is 90°.
So, T = 2 I a b B
Right ?
 
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