question_answer

A circular coil of 200 turns and radius 10 cm is placed in a uniform magnetic field of 0.5 T, parallel to the plane of the coil. If the current in the coil is 3 A, calculate the

(i) total torque on the coil.

(ii) total force on the coil.

(iii) average force on each electron in the coil, due to the magnetic field.

Assume the area of cross-section of the wire to be \[{{10}^{-5}}{{m}^{2}}\] and the free electron density is \[{{10}^{29}}{{m}^{-3}}.\]

Answer:

(i) \[9.42\,N-m\]  (ii) Zero  (iii) \[1.5\times {{10}^{-24}}N\] Given, \[N=200,\]\[r=10\,cm=0.1\,m,\]\[B=0.5\,T\] \[\therefore \]      \[\theta =90{}^\circ ,\] also \[I=3\,A\] (i) As,    \[\tau =NIAB\]             \[=200\times 3\times [\pi \,{{(0.1)}^{2}}]\times 0.5\]  \[[\because \,\,A=\pi {{r}^{2}}]\] \[\Rightarrow \]   \[\tau =9.42\,N\text{-}m\] (ii) The net magnetic force on circular loop is zero. (iii) \[\therefore \] Average force on electron, \[F=(-\,e)({{v}_{d}})\,B\sin 90{}^\circ \] But, \[I=ne\,\,\,A{{\text{v}}_{\text{d}}}\] [where, A = cross-section of the wire]                         \[{{\text{v}}_{\text{d}}}=\frac{I}{-\,neA}\]             \[\therefore \]      \[F=(-\,e)\left( \frac{1}{neA} \right)B\]                         \[F=\frac{IB}{nA}=\frac{3\times 0.5}{{{10}^{29}}\times {{10}^{-5}}}\]             \[\Rightarrow \]   \[F=\frac{1.5}{{{10}^{24}}}\]             \[\Rightarrow \]   \[F=1.5\times {{10}^{24}}N\]