Figure shows a current-carrying square loop ABCD of edge length ‘a’ lying in a plane. If the resistance of the path ABC is r and the path ADC is 2r, find the magnitude of the resultant magnetic field at the centre of the square loop.

 

🧲 Magnetic Field at Centre of Square Loop with Unequal Resistance | JEE Concept

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📌 Question:

Figure shows a current-carrying square loop ABCD of edge length ‘a’ lying in a plane. If the resistance of the path ABC is r and the path ADC is 2r, find the magnitude of the resultant magnetic field at the centre of the square loop.

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🖼️ Question Image:

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💡 Concept Used:

• Unequal resistance in different paths means current divides unequally in branches.

• Use current division rule and Biot-Savart Law or symmetry to find net magnetic field at center.

• Magnetic field due to a straight segment at center of square is:

  $$B=\frac{{\mu }_{0}I}{4\pi R}\cdot 2\sin (\theta \mathrm{/}2)$$

  or use full loop formula if current is symmetrical.

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🧠 Approach:

1. Let total current be I.

2. Since resistances are in parallel:

   • Current in ABC: $I_1 = \frac{2}{3}I$

   • Current in ADC: $I_2 = \frac{1}{3}I$

3. Both paths form half-loops and contribute to B at center.

4. Direction of magnetic fields due to both currents is same (by right-hand thumb rule).

5. Field at center due to full square of current I is:

   $$B=\frac{2\sqrt{2}{\mu }_{0}I}{\pi a}$$

6. Using weighted contribution:

   $$B_{net}=B_{ABC}+B_{ADC}=\frac{2\sqrt{2}{\mu }_0}{\pi a}\left(\frac{2I}{3}\right)$$

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🖼️ Solution Image:

📺 Watch the Full Video Solution:

🧠 Pro Tip:

In problems with non-uniform current distribution, always apply Kirchhoff’s Laws or Current Division first. Then calculate individual contributions to the magnetic field using Biot-Savart or Ampere’s Law.

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