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Torque experienced by a current loop placed in a uniform magnetic field


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Posted

Revered Members,

In my attachment, while measuring the angle between current element I(QR) and magnetic field B, they are going in clock wise direction that is from R to Q(shown in Red line), so it is (90 - Θ ) and while measuring the angle between current element I(SP) and B, they are going in anti clockwise direction that is from P to S(shown in Red line), so it is (90 + Θ).

May I get clarification from revered members as to why they have gone in different directions to measure the angle, instead of adopting the same direction, either clockwise or anti clockwise, for measuring the angle in the arms QR and SP?

My friend told, the reason is,Because they measure the angle according to the direction of current.

My query is

The current starts from the point P and terminates at Q(for the arm PQ) and similarly it starts from point S and terminates at P(for the arm SP), so what harm in measuring the angle from the starting point rather the terminating point of current in both the arms(PQ and SP)?

post-58372-0-06683600-1319334342_thumb.jpg

Posted

Because you are using the cross product. F = I x B so you have to use the directions and angle that are defined by the equation. SP and QR are in opposite directions, so the different angles are defined by the cross product.

 

(etiquette tip: coloring entire passages, especially in that awful green, makes it hard to read. Highlighting is for words or short phrases)

Posted

Thanks a lot swansont. Agreed, i am using cross product. My question is, why can't we measure the angle from Q to R in the arm QR and from S to P in the arm SP, instead of going the way as mentioned in my attachment?

Posted

Thanks a lot swansont. Agreed, i am using cross product. My question is, why can't we measure the angle from Q to R in the arm QR and from S to P in the arm SP, instead of going the way as mentioned in my attachment?

 

You do measure that way. QR and SP. But QR is going to look like PS so the angle you use starts on the other side of the B vector.

Posted

You have a diagram showing one of the segments. The other segment looks exactly the same, except the current flow is in the opposite direction. That forces you to use the opposite angle.

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