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 $\mu_0=4\pi\times10^{-7}\,\text{NA}^{-2}$)
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For a rectangular loop with sides $a$ and $b$, the magnetic field at the center is $B = \frac{2\mu_0 I}{\pi} \frac{\sqrt{a^2+b^2}}{ab}$. This formula is a massive time-saver for competitive exams.
Concept:
The magnetic field at a distance $r$ from a finite straight wire carrying current $I$ is:
\[ B = \frac{\mu_0 I}{4\pi r} (\sin \theta_1 + \sin \theta_2) \]
At the center of a rectangle, the total field is the sum of fields from all 4 sides. Due to symmetry and anti-clockwise current, all fields point in the $+z$ direction.
Step 1: Calculate field from the two longer sides (length $L = 4\sqrt{2}$).
Distance from center $r_1 = 4/2 = 2$ m. Half-length is $2\sqrt{2}$.
$\tan \theta = \frac{2\sqrt{2}}{2} = \sqrt{2} \implies \sin \theta = \frac{\sqrt{2}}{\sqrt{(\sqrt{2})^2 + 1^2}} = \frac{\sqrt{2}}{\sqrt{3}}$.
\[ B_L = 2 \times \left[ \frac{\mu_0 I}{4\pi (2)} \left( 2 \times \frac{\sqrt{2}}{\sqrt{3}} \right) \right] = \frac{\mu_0 I}{2\pi} \frac{\sqrt{2}}{\sqrt{3}} \]
Step 2: Calculate field from the two shorter sides (breadth $W = 4$).
Distance from center $r_2 = 4\sqrt{2}/2 = 2\sqrt{2}$ m. Half-breadth is 2.
$\tan \phi = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}} \implies \sin \phi = \frac{1}{\sqrt{(1)^2 + (\sqrt{2})^2}} = \frac{1}{\sqrt{3}}$.
\[ B_W = 2 \times \left[ \frac{\mu_0 I}{4\pi (2\sqrt{2})} \left( 2 \times \frac{1}{\sqrt{3}} \right) \right] = \frac{\mu_0 I}{2\pi \sqrt{2}} \frac{1}{\sqrt{3}} \]
Step 3: Total Field calculation.
Using the standard formula for a rectangular loop with sides $a$ and $b$:
\[ B = \frac{2\mu_0 I}{\pi} \frac{\sqrt{a^2+b^2}}{ab} \]
Substituting $a = 4\sqrt{2}$, $b = 4$, $I = 5$ A, and $\mu_0 = 4\pi \times 10^{-7}$:
\[ B = \frac{2 (4\pi \times 10^{-7}) (5)}{\pi} \frac{\sqrt{(4\sqrt{2})^2+4^2}}{(4\sqrt{2})(4)} \]
\[ B = 40 \times 10^{-7} \frac{\sqrt{32+16}}{16\sqrt{2}} = 40 \times 10^{-7} \frac{\sqrt{48}}{16\sqrt{2}} \]
\[ B = 40 \times 10^{-7} \frac{4\sqrt{3}}{16\sqrt{2}} = \frac{10\sqrt{3} \times 10^{-7}}{\sqrt{2}} \approx 1.22 \times 10^{-5} \text{ T} \]
