InterviewSolution
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InterviewSolution
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A uniform magnetic field of `3000G` is established along the positive z-direction. A rectangular loop of sides `10cm` and `5cm` carries a current `12A`. What is the torque on the loop in the different cases shown in the figure. What is the force on each case? Which case corresponds to stable equilibrium? |
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Answer» Here, `B=3000G=3000xx10^-4T=0*3T, A=10xx5=50cm^2=50xx10^-4m^2,I=12A` `:. IA=12xx50xx10^-4=0*60Am^2` (a) Here, `IvecA=0*06hatiAm^2`, `vecB=0*3hatkT` Now, torque, `vectau=vec(IA)xxvec(B)=0*06hatixx0*3hatk=-1*8xx10^-2hatjNm` i.e., Torque`=1*8xx10^-2Nm`. It acts along the negative y direction. (b) Here, `vec(IA)=0*06hatiAm^2`, `B=0*3hatkT` Torque, `tau=vec(IA)xxvec(B)=0*06hatixx0*3hatk=-1*8xx10^-2hatjNm` i.e. Torque `=1*8xx10^-2Nm`. It acts along the negative y-direction. (c) Here, `vec(IA)=-0*06hatjAm^2, vecB=0*3hatkT` Torque, `vec(IA)xxvecB=0*06hatjxx0*3hatk=-1*8xx10^-2hatiNm` i.e., Torque `=1*8xx10^-2Nm`. It acts along the negative x-direction. (d) Here, `IA=0*06Am^2`, `B=0*3` Torque, `|vectau|=IAB=0*06xx0*3T=1*8xx10^-2Nm` Here, direction of torque is `30^@+90^@=120^@` anticlockwise with negative x direction i.e. `240^@` with positive x-direction. (e) Here, `vec(IA)=0*06hatkAm^2, vecB=0*3hatkT` Torque, `vectau=IvecAxxvecB=0*06hatkxx0*3hatk=0` (f) Here, `vec(IA)=-0*06hatkAm^2, vecB=0*3hatkT` Torque, `vectau=IvecAxxvecB=-0*06hatkxx0*3hatk=0` The resultant force is zero in all cases. In case (e), `vecIA` and `vecB` are in the same direction, i.e., angle between them `theta=0^@`. Its equilibrium is stable because if loop is distributed a little from this position, it will restore its initial positio. In case (f), `vec(IA)` and `vecB` are in the opposite direction, i.e., angle between them `theta=180^@`. Its equilibrium is unstable because if loop is disturbed a little from this position, it will not restore to its initial position. |
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