Magnetic Field at Centre of Square Loop Problem

❓ Question

Figure shows a current carrying square loop $ABCD$ of side length $r$ lying in a plane.

If the resistance of side $ABCD$ of arc $ADC$ part is:

$$2r:1r$$

then the magnitude of the resultant magnetic field at centre of the square loop is:

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🖼 Question Image

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✍️ Short Explanation

This problem is based on:

👉 Magnetic field due to current carrying wire
👉 Parallel current division
👉 Field at centre of square loop.

Main idea:

Current divides inversely proportional to resistance.

Magnetic field due to a side of square:

$$B=\frac{\mu_0 I}{2\pi r}(\sin\theta_1+\sin\theta_2)$$

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🔷 Step 1 — Current Division 💯

Given resistance ratio:

$$R_{ABC}:R_{ADC}=2:1$$

Current divides inversely:

$$I_{ABC}:I_{ADC}=1:2$$

Let total current be:

$$I$$

Then:

$$I_{ABC}=\frac{I}{3}$$
$$I_{ADC}=\frac{2I}{3}$$

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🔷 Step 2 — Geometry of Square

Side length:

$$r$$

Distance of centre from each side:

$$\frac{r}{2}$$

For one side of square:

$$\theta_1=\theta_2=45^\circ$$

Magnetic field due to one side:

$$B_{side} = \frac{\mu_0 I}{4\pi(r/2)} (\sin45^\circ+\sin45^\circ)$$
$$= \frac{\mu_0 I}{2\pi r} \left(\frac1{\sqrt2}+\frac1{\sqrt2}\right)$$
$$= \frac{\mu_0 I}{\sqrt2\pi r}$$

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🔷 Step 3 — Field Due to Upper Path

Path $ABC$ contains two sides.

Current through it:

$$\frac{I}{3}$$

Hence:

$$B_1 = 2\left( \frac{\mu_0}{\sqrt2\pi r} \cdot\frac{I}{3} \right)$$
$$= \frac{\sqrt2\mu_0 I}{3\pi r}$$

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🔷 Step 4 — Field Due to Lower Path

Path $ADC$ also contains two sides.

Current through it:

$$\frac{2I}{3}$$

Thus:

$$B_2 = 2\left( \frac{\mu_0}{\sqrt2\pi r} \cdot\frac{2I}{3} \right)$$
$$= \frac{2\sqrt2\mu_0 I}{3\pi r}$$

Direction of both fields at centre is opposite.

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🔷 Step 5 — Resultant Magnetic Field

$$B=B_2-B_1$$
$$= \frac{2\sqrt2\mu_0 I}{3\pi r} - \frac{\sqrt2\mu_0 I}{3\pi r}$$
$$= \frac{\sqrt2\mu_0 I}{3\pi r}$$

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🔷 Step 6 — Final Simplification

$$\boxed{ B=\frac{\mu_0 I}{3\sqrt2\pi r} }$$

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🔷 Step 7 — JEE Trap Alert 🚨

❌ Current equally divide assume kar lena

❌ Magnetic field directions ignore kar dena

❌ Distance from side to centre $r$ le lena instead of $r/2$

Remember:

$$\boxed{ I\propto\frac1R }$$

for parallel branches.

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✅ Final Answer

$$\boxed{ \frac{\mu_0 I}{3\sqrt2\pi r} }$$

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📚 Related Topics

• Electric Flux Rate and Displacement Current
• Current Calculation with Zener Breakdown
• Displacement Current from Changing Electric Flux