Question

A magnetic pole of strength \[4 Am\] is moved twice around a long straight wire carrying a current of \[3A\]. The work done is-
A.\[96\pi \times {10^{ - 7}}J\]
B.\[48 \times {10^{ - 7}}J\]
C.\[108 \times {10^{ - 7}}J\]
D.\[zero\]

Answer 1

Hint: After applying Biot-Savart’s law by bearing in mind an elementary length on the finite straight wire. For any long or infinite length of the straight wire or for any conductor, the perpendicular distance from the wire is next to the center of the wire. The direction of the magnetic field is constantly in a plane which is perpendicular to the line of element and position vector

 Complete answer:
The direction of the magnetic field \[B\] is alongside the tangent of the circular wire.
The radius of the circle is given by \[r\]
Carrying current is \[i = {\text{ }}3A\]
The formula of the magnetic field according to Biot-Savart’s law, \[B = \dfrac{{\left( {{\mu _0} \times i} \right)}}{{\left( {2 \times \pi \times r} \right)}}\]
So by putting the value of the current, we get \[B = \dfrac{{\left( {{\mu _0} \times 3} \right)}}{{\left( {2 \times \pi \times r} \right)}} = \dfrac{{3{\mu _0}}}{{2\pi r}}\]
Therefore the Force \[F = mB = \dfrac{{3m{\mu _0}}}{{2\pi r}}\]
Now the work done is given by the following equation
\[W = F \times 2\left( {2\pi r} \right)\]
\[ \Rightarrow W = \dfrac{{3m{\mu _0}}}{{2\pi r}} \times 4\pi r\]
\[ = 6{\mu _0}\pi m\]
\[ = 6 \times 4 \times {10^{^ - 7}} \times \pi \times 4\]
\[ = 96\pi \times {10^{^ - 7}}J\]
Hence, we get the option (A) as the correct answer.
But, according to Gauss's law, a monopole does not exist.
So, considering the hypothetical case since the magnetic field is generated in the circular loop by the current-carrying conductor, work done will be zero. Then the correct answer will be option (D).

Hence, we get the option (A) as the correct answer.

 Note:
Biot-Savart law is the same as Coulomb’s law in electrostatics. It is specified by the right-hand thumb rule where the thumb points to the path of conventional current and the remaining fingers indicate the magnetic field’s direction. Let us consider the magnetic field for 3 different cases.
 1) When the wire has a finite length, we get $B = \dfrac{{{\mu _0}I}}{{4\pi a}}\left( {\sin {\phi _2} + \sin {\phi _1}} \right)$.
2) When the wire has infinite length, we get ${\phi _1} = {\phi _2} = {90^\circ }$. Which implies, $B = \dfrac{{{\mu _0}I}}{{2\pi a}}$
 3) When the wire has infinite length and the point $P$ lies near the wire’s end, ${\phi _1} = {90^\circ },{\phi _2} = 0\,\,$
Which implies, $B = \dfrac{{{\mu _0}I}}{{4\pi a}}$