Question

If the point $\text{P}$ lies outside the line of straight finite wire $\text{AB}$, as shown in the figure, then the magnitude of magnetic field at point $\text{P}$ is:

• A. $\frac{{\mu }_{0}I}{4\pi d}(\sin {\alpha }_1+\sin {\alpha }_2)$
• B. $\frac{{\mu }_{0}I}{4\pi d}(\cos {\alpha }_1+\cos {\alpha }_2)$
• C. $\frac{{\mu }_{0}I}{4\pi d}(\cos {\alpha }_1-\cos {\alpha }_2)$
• D. $\frac{{\mu }_{0}I}{4\pi d}(\sin {\alpha }_1-\sin {\alpha }_2)$

Answer 1

The correct option is C $\frac{{\mu }_{0}I}{4\pi d}(\cos {\alpha }_1-\cos {\alpha }_2)$

The perpendicular distance of $P$ from wire is $d$. Using the realtion for magnetic field due to straight finite wire,
$$B_p=\frac{{\mu }_{0}I}{4\pi d}(\sin {\varphi }_1+\sin {\varphi }_2)...\left(1\right)$$Here, we have to look for angles ${\varphi }_1$ and ${\varphi }_2$,


From geometry, ${\varphi }_1=\frac{\pi }{2}-{\alpha }_1$ and ${\varphi }_2=\frac{\pi }{2}-{\alpha }_2$

The angle ${\varphi }_1$ is angle substended below the perpendicular line $OP$ and angle ${\varphi }_2$ should be the angle substended by the others end of the wire above line $OP$.
Thus, ${\varphi }_2=-(\frac{\pi }{2}-{\alpha }_2)$

Using equation $\left(1\right)$,

$B_p=\frac{{\mu }_{0}I}{4\pi d}[\sin (\frac{\pi }{2}-{\alpha }_1)+\sin [-(\frac{\pi }{2}-{\alpha }_1)]]$

$\Rightarrow B_P=\frac{{\mu }_{0}I}{4\pi d}[\cos {\alpha }_1-\sin (\frac{\pi }{2}-{\alpha }_1)]$

$\Rightarrow B_P=\frac{{\mu }_{0}I}{4\pi d}[\cos {\alpha }_1-\cos {\alpha }_2]$

Hence, option (c) is correct answer.