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Derive an expression for the torque acting on a current carrying loop suspended in a uniform magnetic field. |
Answer» Solution :1. As shown in FIGURE, consider a rectangular coil ABCD suspended in a uniform magnetic field `vecB`, with its AXIS perpendicular to the field.   2. Suppose, current flowing through coil ABCD be I, AB = DC = b and AD = BC = a and area be A = ab. 2. The field exerts no force on the two arms AD and BC of the loop, because `I_(a)andvecB` both are parallel to each other, `|vecF|=|IvecaxxvecB|` `|vecF|=(IaxxB)SIN0^(@)=0""...(1)` 3. Sides AB and CD of coils are perpendicular to magnetic field. 4. So, force on side AB is, `vecF_(1)=|IvecbxxvecB|=Ib"sin"pi/2` `|vecF_(1)|=IbB""...(2)` 5. Similarly it exerts a force `vecF_(2)` on the arm CD and `vecF_(2)` is directed out of the plane of the paper, `|vecF_(2)|=|IvecbxxvecB|=|Ib"sin"pi/2|` `thereforeF_(2)=IbB=F_(1)""...(2)` 6. Thus, the net force on the loop is zero. There is a torque on the loop due to the pair of forces `vecF_(1)andvecF_(2)`. 7. Figure (b) shows a view of the loop from the AD end. It shows that the torque on the loop tends to rotate it anti-clockwise. 8. Resultant torque on coil is given by, `tau=tau_(1)+tau_(2)` `tau=F_(1)(a/2)+F_(2)(a/2)` = `IbB(a/2)+IbB(a/2)` = `I(ab)B` `tau=IAB""...(3)` where A = ab that is area of RECTANGLE. |
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