question_answer

A long copper wire carries a steady current. Force on an electron moving with drift speed on surface of wire will be take current in wire \[I=10A,\] radius of wire \[a=0.5\,mm\] and number of conduction electrons in wire \[n=5\times {{10}^{28}}{{m}^{-\,3}}.\]

A) \[\pi \times {{10}^{-\,25}}N\]

B) \[2\pi \times {{10}^{-\,26}}N\]

C) \[4\pi \times {{10}^{-\,25}}N\]

D) \[8\pi \times {{10}^{-\,26}}N\]

Correct Answer: D

Solution :

Direction of force on an electron using Flemming's right hand rule is towards centre of wire.

Force on electron is \[F=Bq{{v}_{d}}\]

where, \[{{v}_{d}}=\frac{I}{nqA}\cdot \]

\[\therefore \]      \[F=\frac{{{\mu }_{0}}I}{2\pi a}\cdot q\cdot \frac{I}{nq\pi {{a}^{2}}}\]

\[F=\frac{{{\mu }_{0}}{{I}^{2}}}{2{{\pi }^{2}}n{{a}^{3}}}\]

Substituting values in above equation, we have  \[F=\frac{4\times \pi \times {{10}^{-\,7}}\times {{10}^{2}}}{2\times {{\pi }^{2}}\times 5\times {{10}^{28}}\times {{(0.5\times {{10}^{-\,3}})}^{3}}}\]

\[\approx 8\pi \times {{10}^{-\,26}}N\]