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For the magnetic field to be maximum due to small element of current carrying conductor at a point, the angle between the element and the line joining the element to the given point must be A. ${0^ \circ }$B. ${90^ \circ }$C. ${180^ \circ }$D. ${45^ \circ }$

Last updated date: 11th Sep 2024
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Hint: Charges at rest produce only electric fields while charges in motion produce both electric and magnetic fields. Current is flowing through wire means that charge is in motion which produces the magnetic field and we will calculate the magnetic field produced by it at the required point
Formula used:
$\mathop {dB}\limits^ \to = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{I(\mathop {dl}\limits^ \to \times \mathop r\limits^ \to )}}{{{{\left| r \right|}^3}}}$

Electricity and magnetism are interdependent on each other. Motion of the magnet in the conducting loop generates the eddy currents while motion of charge in the conductor produces the magnetic field around it.
In case of current carrying conductor we will consider a small elemental length from the conductor and magnetic field is found out due to that elemental length and we will integrate it throughout the length of the conductor to get the magnetic field at the required point.
Magnetic field intensity due to small elemental length $\mathop {dl}\limits^ \to$ from the conductor in which ‘I’ current is passing and the distance between that point and element is ‘r’ is $\mathop {dB}\limits^ \to$
So that magnetic field is given by the formula
$\mathop {dB}\limits^ \to = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{I(\mathop {dl}\limits^ \to \times \mathop r\limits^ \to )}}{{{{\left| r \right|}^3}}}$
If we consider only magnitude of the above vector we will get
$\mathop {dB}\limits^ \to = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{I(\mathop {dl}\limits^ \to \times \mathop r\limits^ \to )}}{{{{\left| r \right|}^3}}}$
$\therefore \left| {\mathop {dB}\limits^ \to } \right| = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{I\left| {\mathop {dl}\limits^ \to } \right|\sin \theta }}{{\left| r \right|2}}$
The above expression will be maximum when $\sin \theta$ is maximum and maximum value of $\sin \theta$ is one.
\eqalign{ & \sin \theta = 1 \cr & \therefore \theta = {90^ \circ } \cr}

Hence option B is the required answer.

Note:
If we want to find the magnetic field due to long straight current carrying conductor then we will take small elemental length and integrate over total length by following this method. There is another way. We can apply ampere circuital law to find out the magnetic field at a point. Both yield the same result.