Question

A rectangular conducting loop of length $4\sqrt{2}m$ and breadth 4 m carrying a current of 5 A in the anti-clockwise direction is placed in the xy-plane. The magnitude of the magnetic induction field vector B at the intersection of the diagonals is
(Use ${\mu }_0=4\pi \times 10^{-7}N{A}^{-2}$)

• A. $1.2\times 10^{-6}T$
• B. $1.2\times 10^{-5}T$
• C. $2.4\times 10^{-6}T$
• D. $2.4\times 10^{-5}T$
• E. $1.2\times 10^{-7}T$

Answer 1

Answer: (A)

Magnetic field due to a straight current carrying conductor of finite length
$B=\frac{{\mu }_0}{4\pi }\frac{I}{d}\left(\sin {\theta }_1+\sin {\theta }_2\right)...(i)$
(i) Magnetic field due to conductor $DA$.
Here, $d=\frac{4}{2}=2m$
${\theta }_1={\theta }_2={\tan }^{-1}\left(\frac{2\sqrt{2}}{2}\right)=54.73^{\circ }$
$\therefore \sin {\theta }_1=\sin {\theta }_2=\sin 54.73^{\circ }=0.816$
and $I=5A$ (given)
From Eq. (i),
$B_1=\frac{{\mu }_0}{4\pi }\times \frac{5}{2}(0.816+0.816)$
$B_1=4.08\times 10^{-7}T$
Similarly, magnetic field due to conductor $BC$
$B_2=4.08\times 10^{-7}T$
(ii) Magnetic field due to conductor $AB$.
Here, $d=\frac{4\sqrt{2}}{2}=2\sqrt{2}m$
${\theta }_1={\theta }_2={\tan }^{-1}\left(\frac{2}{2\sqrt{2}}\right)=35.26^{\circ }$
$\therefore \sin {\theta }_1=\sin {\theta }_2=\sin 35.26^{\circ }=0.577$
and $I=5A$ (given)
From Eq. (i),
$B_3=\frac{{\mu }_0}{4\pi }\times \frac{5}{2\sqrt{2}}(0.57+0.57)$
$B_3=2.04\times 10^{-7}T$
Similarly magnetic field due to conductor $CD$
$B_4=2.04\times 10^{-7}T$
Hence, magnitude of induction field vector $B$ at the intersection of the diagonals
$B=B_1+B_2+B_3+B_4$
$B=(4.08+4.08+2.04+2.04)\times 10^{-7}$
$B=12.24\times 10^{-7}T$
$B=1.2\times 10^{-6}T$