Question:
A current of 5 A exists in a square loop of side length 5 cm. The magnitude of the magnetic field at the centre of the square loop is approximately:
A current of 5 A exists in a square loop of side length 5 cm. The magnitude of the magnetic field at the centre of the square loop is approximately:
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At the centre of a square loop, all four sides contribute equally and add in the same direction.
Updated On: Jun 20, 2026
- 3.8 mT
- 1.7 mT
- 4.8 mT
- 2.3 mT
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The Correct Option is D
Solution and Explanation
Step 1: Concept of magnetic field due to current carrying wire.
Magnetic field at a point due to a finite straight wire is: \[ B = \frac{\mu_0 I}{4\pi r}(\sin\theta_1 + \sin\theta_2) \]
Step 2: Field due to one side of square.
At the centre of a square, each side subtends equal angles: \[ \theta_1 = \theta_2 = 45^\circ \] So, \[ \sin 45^\circ + \sin 45^\circ = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} \] Distance from centre to side: \[ r = \frac{a}{2} \]
Step 3: Magnetic field due to one side.
\[ B_1 = \frac{\mu_0 I}{4\pi (a/2)} \cdot \sqrt{2} \] \[ B_1 = \frac{\mu_0 I \sqrt{2}}{2\pi a} \]
Step 4: Total field due to 4 sides.
All four sides contribute equally and in same direction: \[ B = 4B_1 = \frac{2\sqrt{2}\mu_0 I}{\pi a} \]
Step 5: Substitute values.
\[ I = 5 \, A,\quad a = 5 \, cm = 0.05 \, m,\quad \mu_0 = 4\pi \times 10^{-7} \] \[ B = \frac{2\sqrt{2} \cdot 4\pi \times 10^{-7} \cdot 5}{\pi \cdot 0.05} \] Cancel \(\pi\): \[ B = \frac{2\sqrt{2} \cdot 20 \times 10^{-7}}{0.05} \] \[ B = \frac{40\sqrt{2} \times 10^{-7}}{0.05} \]
Step 6: Final calculation.
\[ B \approx 2.3 \times 10^{-3} \, T \] \[ B \approx 2.3 \, mT \]
Step 7: Final conclusion.
\[ \boxed{2.3 \, mT} \]
Magnetic field at a point due to a finite straight wire is: \[ B = \frac{\mu_0 I}{4\pi r}(\sin\theta_1 + \sin\theta_2) \]
Step 2: Field due to one side of square.
At the centre of a square, each side subtends equal angles: \[ \theta_1 = \theta_2 = 45^\circ \] So, \[ \sin 45^\circ + \sin 45^\circ = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} \] Distance from centre to side: \[ r = \frac{a}{2} \]
Step 3: Magnetic field due to one side.
\[ B_1 = \frac{\mu_0 I}{4\pi (a/2)} \cdot \sqrt{2} \] \[ B_1 = \frac{\mu_0 I \sqrt{2}}{2\pi a} \]
Step 4: Total field due to 4 sides.
All four sides contribute equally and in same direction: \[ B = 4B_1 = \frac{2\sqrt{2}\mu_0 I}{\pi a} \]
Step 5: Substitute values.
\[ I = 5 \, A,\quad a = 5 \, cm = 0.05 \, m,\quad \mu_0 = 4\pi \times 10^{-7} \] \[ B = \frac{2\sqrt{2} \cdot 4\pi \times 10^{-7} \cdot 5}{\pi \cdot 0.05} \] Cancel \(\pi\): \[ B = \frac{2\sqrt{2} \cdot 20 \times 10^{-7}}{0.05} \] \[ B = \frac{40\sqrt{2} \times 10^{-7}}{0.05} \]
Step 6: Final calculation.
\[ B \approx 2.3 \times 10^{-3} \, T \] \[ B \approx 2.3 \, mT \]
Step 7: Final conclusion.
\[ \boxed{2.3 \, mT} \]
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