Question

As shown in the figure, two infinitely long, identical wires are bent by ${90^ \circ }$ and placed in such a way that the segments $LP$ and $QM$ are along the x-axis, while segments $PS$ and $QN$ are parallel to the y-axis, if $OP = OQ = 4cm$, and the magnitude of the magnetic field at $O$ is \[{10^{ - 4}}T\], and the two wires carry equal currents (see figure), the magnitude of the current wire and the direction of the magnetic field at $O$ will be (${\mu _0} = 4\pi {10^{ - 7}}N{A^{ - 2}}$)

A) $40A$, Perpendicular into the page
B) $40A$, perpendicular out of the page
C) $20A$, perpendicular out of the page
D) $20A$, perpendicular into the page

Answer 1

Hint: Recall the formula for the magnetic field due to a straight wire. Always remember that whenever there are two segments along the same axis and the direction of current is opposite to each other then the magnetic field due to these segments is zero. Here only find the magnetic field due to the segment $PS$ and $QN$.

Formula used:
Magnetic field due to a straight wire,

$B = \dfrac{{{\mu _0}i}}{{4\pi r}}\left( {\sin \alpha + \sin \beta } \right)$
Here, ${\mu _0}$ is the permeability of free space
$i$ is the current flowing in the wire
$r$ is the distance of the point to the wire, at which the magnetic field is calculated
$\alpha $ and $\beta $ are the angles formed with that point

Complete step by step solution:
As we know that the segment $LP$ and $QM$ are along the x-axis. Therefore, they formed an angle of ${0^ \circ }$ with the point $O$.
So, according to the formula of magnetic field due to a straight wire,
${B_{LP}} = {B_{MQ}} = 0$
Now, the magnetic field due to the segment $PS$
Segment $PS$ is a half wire and parallel to the y-axis,
Therefore, $\alpha = {90^ \circ }$ and $\beta = {0^ \circ }$
Putting these values in the formula for the magnetic field due to straight wire,
${B_{PS}} = \dfrac{{{\mu _0}i}}{{4\pi r}}\left( {\sin {{90}^ \circ } + \sin {0^ \circ }} \right)$
We know that, $\sin {90^ \circ } = 1$
$\sin {0^ \circ } = 0$
On putting the values in the above equation, we get
${B_{PS}} = \dfrac{{{\mu _0}i}}{{4\pi r}} \otimes $ {direction is into the plane}
Now, the magnetic field due to the segment $QN$

Since, both the segment $PS$ and $QN$ are equal but the direction is opposite. So, the magnitude of the magnetic field due to the segment $QN$ is equal to the magnetic field due to the segment $PS$but the direction is also into the plane.
${B_{QN}} = \dfrac{{{\mu _0}i}}{{4\pi r}} \otimes $ {direction is into the plane}
Now, the magnetic field at point $O$ is given by,
${B_O} = {B_{PS}} + {B_{QN}} + {B_{LP}} + {B_{MQ}}$
On putting all the values in the above equation , we get
${B_O} = \dfrac{{{\mu _0}i}}{{4\pi r}} + \dfrac{{{\mu _0}i}}{{4\pi r}} + 0 + 0$
As the direction of both ${B_{PS}}$ and ${B_{QN}}$ is inward. So, we simply add them
$ \Rightarrow {B_O} = 2 \times \dfrac{{{\mu _0}i}}{{4\pi r}}$ …………..(i)
So, in the question we have given the magnitude of the magnetic field at $O$,
${B_O} = {10^{ - 4}}T$
And also given the distance $OP = OQ = 4cm$, which is,
$r = 4 \times {10^{ - 2}}m$
We know that, $\dfrac{{{\mu _0}}}{{4\pi }} = {10^{ - 7}}N{A^{ - 2}}$
On putting all the values in equation (i), we get
$ \Rightarrow {10^{ - 4}} = \dfrac{{2 \times {{10}^{ - 7}} \times i}}{{4 \times {{10}^{ - 2}}}}$
On further solving, we get the magnitude of the current,
$i = 20A$
Thus, we get the magnitude of the current is $20A$ and the direction of the magnetic field at $O$ will be perpendicular to the page.

Therefore, the correct answer is option (D).

Note: The direction of the magnetic field in the wire is found with the right hand thumb rule. If you hold the wire with your right hand so that your thumb points along the current, then your fingers wrap around the wire in the same sense as the magnetic field direction.