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 $\mu_0 = 4 \pi \times 10^{-7} \, NA^{-2}$)

• 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 $D A$.
Here, $d=\frac{4}{2}=2 m$
$\theta_{1}=\theta_{2}=\tan ^{-1}\left(\frac{2 \sqrt{2}}{2}\right)=54.73^{\circ}$
$\therefore \sin \theta_{1}=\sin \theta_{2}=\sin 54.73^{\circ}=0.816$
and $I=5 A$ (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 $B C$
$B_{2}=4.08 \times 10^{-7} T$
(ii) Magnetic field due to conductor $A B$.
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 $I=5 A$ (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 $C D$
$B_{4}=2.04 \times 10^{-7} T$
Hence, magnitude of induction field vector $B$ at the intersection of the diagonals
$B=B_{1}+B_{2}+B_{3}+B_{4}$
$B=(4.08+4.08+2.04+2.04) \times 10^{-7}$
$B=12.24 \times 10^{-7} T$
$B=1.2 \times 10^{-6} T$