InterviewSolution
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InterviewSolution
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Derive the expression for the torque tau acting ona rectangular current loop of area A placed in a uniform magnetic field B. Show thatvec tau=vecmxx vec B where vecm is the moment of the current loop given by vecm =vecIA. |
Answer» Solution :Consider a rectangular coil PQRSsuspended in a uniform magnetic field `vecB`, with its axis perpendicular to the field.  Let `I=` current flowing through the coil PQRS `a,b=` sides of the coil PQRS ` A= ab= `AREA of the coil. ` theta =` angle between the direction of `vecB` and normal to the plane of the coil. According to Fleming's left hand rule, the magnetic forces on sides PS and QR are equal, OPPOSITE and collinear (along the axis of the loop), so their resultant is zero. The side PQ experiences a normal inward force equal to lbB while the side RS experiences an equal normal outward force. These two forces form a couple which exerts a troque given by. ` tau=` force x perpendicular distance `=lbBxx asin theta =lBA sin theta` If the rectangular loop has N turns, the torque increases N times i.e., `vec tau =NIBA sin theta` But `NIA =m`, the magnetic moment of the loop , so ` tau =mB sin theta` In vector NOTATION, the torque`vectau` is given by ` vectau=vecmxx vecB`. |
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