Question

Find the magnetic field at point $P$ due to a straight wire segment $AB$ of length $6\text{cm}$ carrying a current of $5\text{A}$.

• A. $2.0\times {10}^{-5}\text{T}$
• B. $1.5\times {10}^{-5}\text{T}$
• C. $3.0\times {10}^{-5}\text{T}$
• D. $2.5\times {10}^{-5}\text{T}$

Answer 1

The correct option is B $1.5\times {10}^{-5}\text{T}$


Magnetic field at point $P$,

$B_P=\frac{{\mu }_0}{4\pi }\frac{I}{a}[\sin \theta +\sin \theta ]$

From $\Delta \text{APC}$,

$A{P}^2=A{C}^2+C{P}^2$

$\Rightarrow C{P}^2=A{P}^2-A{C}^2$

$\Rightarrow CP=\sqrt{A{P}^2-A{C}^2}=\sqrt{5^2-3^2}=4\text{cm}$

$\therefore a=CP=4\text{cm}$

$\sin \theta =\frac{AC}{AP}=\frac{3}{5}$

$\Rightarrow B_P={10}^{-7}\times \frac{5}{4\times {10}^{-2}}[\frac{3}{5}+\frac{3}{5}]$

$\therefore B_P=1.5\times {10}^{-5}\text{T}$

Hence, option $\left(b\right)$ is the correct answer.