Magnetic Field on Axis of Circular Coil

 

❓ Question

For a circular coil of radius $R$, the magnetic field at its center is

$$B_0=16\,\mu T.$$

What will be the magnetic field on its axis at a distance

$$x=\sqrt{3}\,R$$

from the center?

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✍️ Short Explanation

The magnetic field on the axis of a circular current-carrying coil at a distance $x$ from its center is

$$B=\frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}.$$

Since the magnetic field at the center is

$$B_0=\frac{\mu_0 I}{2R},$$

we can write

$$B=B_0\left(\frac{R^2}{R^2+x^2}\right)^{3/2}.$$

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🔷 Step 1 — Substitute the Given Values

Given,

$$B_0=16\,\mu T,\qquad x=\sqrt{3}\,R.$$

Hence,

$$x^2=3R^2.$$

So,

$$B=16\left(\frac{R^2}{R^2+3R^2}\right)^{3/2} =16\left(\frac{1}{4}\right)^{3/2}.$$

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🔷 Step 2 — Simplify

$$\left(\frac{1}{4}\right)^{3/2} = \left(\frac{1}{2}\right)^3 = \frac{1}{8}.$$

Therefore,

$$B=16\times\frac{1}{8}=2\,\mu T.$$

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✅ Final Answer

$$\boxed{2\,\mu T}$$

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🔷 JEE Shortcut

Remember the direct relation:

$$\boxed{B=B_0\left(\frac{R^2}{R^2+x^2}\right)^{3/2}}$$

For $x=\sqrt{3}R$,

$$R^2+x^2=4R^2$$

so the multiplying factor becomes

$$\left(\frac{1}{4}\right)^{3/2}=\frac{1}{8}.$$

Thus,

$$\boxed{B=\frac{B_0}{8}=\frac{16}{8}=2\,\mu T.}$$