Question

Current through $ABC$ and $A^{\prime }{B}^{\prime }{C}^{\prime }$ is $I$. What is the magnetic field at $P?BP=P{B}^{\prime }=r$ (Here $C^{\prime }{B}^{\prime }PBC$ are collinear)

• A. $B=\frac{1}{4\pi }\frac{2I}{r}$
• B. $B=\frac{{\mu }_0}{4\pi }\left(\frac{2I}{r}\right)$
• C. $B=\frac{{\mu }_0}{4\pi }\left(\frac{I}{r}\right)$
• D. Zero

Answer 1

Answer: (B)

Magnetic field due the portions $BC$ and $B^{\prime }{C}^{\prime }$ each is zero.
Magnetic field due to $AB$ :
$<br/>B_1=\frac{{\mu }_{0}I}{4\pi (PB)}\left[\sin {\theta }_1+\sin {\theta }_2\right]<br/>$
where, ${\theta }_1=0^{\circ }$ and ${\theta }_2=90^{\circ }$
$<br/>\begin{array}{l}<br/>\Rightarrow B_1=\frac{{\mu }_{0}I}{4\pi r}[0+1]\\ <br/>\Rightarrow B_1=\frac{{\mu }_{0}I}{4\pi r}\text{ (Into the paper) }<br/>\end{array}<br/>$
Magnetic field due to $A^{\prime }{B}^{\prime }$ :
$<br/>B_2=\frac{{\mu }_{0}I}{4\pi \left(P{B}^{\prime }\right)}\left[\sin {\theta }_1^{\prime }+\sin {\theta }_2^{\prime }\right]<br/>$
where, ${\theta }_1^{\prime }=0^{\circ }$ and ${\theta }_2^{\prime }=90^{\circ }$
$<br/>\Rightarrow B_2=\frac{{\mu }_{0}I}{4\pi (r}[0+1]<br/>$
$<br/>\Rightarrow B_2=\frac{{\mu }_{0}I}{4\pi r}\text{ (Into the paper) }<br/>$
Thus, total magnetic field at $P,B_p=B_1+B_2=\frac{{\mu }_o(2I)}{4\pi r}$ (Into the paper)