Question

Find the magnetic field at point P due to a straight line segment $AB$ of length $6cm$ carrying a current of $5A$. (See figure)

$({\mu }_o=4\pi \times {10}^{-7}N{A}^{-2})$
$\text{A}\text{. }3.0\times {10}^{-5}T$
$\text{B}\text{. 2}\text{.5}\times {10}^{-5}T$
$\text{C}\text{. 2}.0\times {10}^{-5}T$
$\text{D}\text{. 1}.5\times {10}^{-5}T$

Answer 1

Hint: To find the value of magnetic field at point P, we will use the formula of magnetic field due to a straight current carrying wire segment at a point present on the equatorial axis of the segment.

Formula used:
$B={\displaystyle \frac{{\mu }_{o}i}{4\pi r}}(\sin {\theta }_1+\sin {\theta }_2)$

Complete step by step answer:
Magnetic field is a vector field, or pseudo vector field, that describes the magnetic influence or impact of electric charges in relative motion with each other and effect of magnetized materials. A charge moving parallel to a current of other charges experiences a force perpendicular to its own velocity. We can say that the magnetic field is the area around a magnet in which there is presence of magnetic force. Magnetic field is a type of field that passes through space and which makes a magnetic force move electric charges and magnetic dipoles.
Suppose MN is a straight conductor carrying a current $I$ and magnetic field intensity is to be determined at point X.

According to Biot-Savart law, magnetic field at point X is,
$\overrightarrow{dB}={\displaystyle \frac{{\mu }_o}{4\pi }}{\displaystyle \frac{I\overrightarrow{dl}\times \overrightarrow{r}}{r^3}}$
Angle between $I\overrightarrow{dl}$ and $\overrightarrow{r}$ is $$(180-\theta )$$, so,
$dB={\displaystyle \frac{{\mu }_o}{4\pi }}{\displaystyle \frac{Idl\sin (180-\theta )}{r^2}}$
$dB={\displaystyle \frac{{\mu }_o}{4\pi }}{\displaystyle \frac{Idl\sin \theta }{r^2}}$
Now, $EG=EF\sin \theta =dl\sin \theta$
And, $EG=EP\sin d\varphi =r\sin d\varphi =rd\varphi$
We get, $dl\sin \theta =rd\varphi$
Therefore, $dB={\displaystyle \frac{{\mu }_o}{4\pi }}{\displaystyle \frac{Id\varphi }{r}}$
Also, $r={\displaystyle \frac{R}{\cos \varphi }}$
$dB={\displaystyle \frac{{\mu }_o}{4\pi }}{\displaystyle \frac{I\cos \varphi d\varphi }{R}}$
Total magnetic field at point X due to entire conductor is,
$$\begin{array}{rl}& B=\underset{-{\varphi }_1}{\overset{{\varphi }_2}{\int }}{\displaystyle \frac{{\mu }_o}{4\pi }}{\displaystyle \frac{1}{R}}\cos \varphi d\varphi \\ & ={\displaystyle \frac{{\mu }_o}{4\pi }}{\displaystyle \frac{I}{R}}\underset{-{\varphi }_1}{\overset{{\varphi }_2}{[\sin \varphi ]}}\end{array}$$
$B={\displaystyle \frac{{\mu }_{o}i}{4\pi r}}(\sin {\theta }_1+\sin {\theta }_2)$
Magnetic field due to a finite current carrying wire is given by,
$B={\displaystyle \frac{{\mu }_{o}i}{4\pi r}}(\sin {\theta }_1+\sin {\theta }_2)$
We are given a wire segment $AB$ of length $6cm$ carrying a current of $5A$ and we have to the find the value of magnetic field at point P
Magnetic field due to a finite current carrying wire is given by,
$B={\displaystyle \frac{{\mu }_{o}i}{4\pi d}}(\sin {\theta }_1+\sin {\theta }_2)$
Distance $d=4cm=4\times {10}^{-2}m$
Current $i=5A$
Angle ${\theta }_1={\theta }_2={37}^{\circ }$
$\sin {\theta }_1=\sin {\theta }_2={\displaystyle \frac{3}{5}}$
${\mu }_o=4\pi \times {10}^{-7}$
Putting all the values, we get,
$B={\displaystyle \frac{5}{4\times {10}^{-2}}}({\displaystyle \frac{3}{5}}+{\displaystyle \frac{3}{5}})\times {10}^{-7}$
$B={\displaystyle \frac{5}{4}}\times 2\times {\displaystyle \frac{3\times {10}^{-7}}{5\times {10}^{-2}}}$
$B=1.5\times {10}^{-5}T$
Value of Magnetic field at point P due to straight line segment $AB$ is $1.5\times {10}^{-5}T$
Hence, the correct option is D.

Note: Students should note that the value of $d$ or $r$ in the formula of Magnetic field is the perpendicular distance from the wire to the point where we have to find the value of the Magnetic field.