Question

A wire carries current I into the junction shown in the figure .
What is the magnetic field at the dot?(B)

Answer 1

Concepts and reason

The concept of right hand thumb rule, current, and magnetic field is required to solve this problem.

Initially, calculate the magnetic field in the upper wire and then calculate the magnetic field in the lower wire by using the expression for magnetic field. Finally, calculate the magnetic field at the dot.

Fundamentals

The expression for magnetic field is,

$B = \frac{{{\mu _0}I}}{{4\pi r}}$

Here, ${\mu _0}$ is the permittivity in free space, I is the current passing through the wire, a is the distance from wire to the dot.

The magnetic field in the upper wire is,

$\begin{array}{c}\\{B_{upper}} = \frac{{{\mu _0}{I_1}}}{{4\pi a}}\cos {\theta _1} + \frac{{{\mu _0}{I_1}}}{{4\pi a}}\cos {\theta _2}\\\\ = \frac{{{\mu _0}\left( {I/2} \right)}}{{4\pi a}}\cos \left( {0^\circ } \right) + \frac{{{\mu _0}\left( {I/2} \right)}}{{4\pi a}}\cos \left( {90^\circ } \right)\\\\ = \frac{{{\mu _0}I}}{{8\pi a}}\\\end{array}$

The magnetic field in the lower wire is will be same as magnetic field in the upper wire but in opposite direction.

${B_{lower}} = - \frac{{{\mu _0}I}}{{8\pi a}}$

The magnetic field at the dot is

$\begin{array}{c}\\{B_{dot}} = {B_{upper}} + {B_{lower}}\\\\ = \frac{{{\mu _0}I}}{{8\pi a}} - \frac{{{\mu _0}I}}{{8\pi a}}\\\\ = 0\\\end{array}$

Ans:

The net magnetic field at the dot is 0.