Question

For a circular coil of radius \(R\), magnetic field at the center is \(B_0 = 16~\mu\text{T}\). What will be the magnetic field on the axis at a distance \(x = \sqrt{3}R\) from the center?

• A. \(\dfrac{1}{4}~\mu\text{T}\)
• B. \(\dfrac{1}{2}~\mu\text{T}\)
• C. \(4~\mu\text{T}\)
• D. \(2~\mu\text{T}\)

Hint:

For magnetic field on the axis of a circular coil:

Always relate the field to the central field \(B_0\).
Substitute \(x/R\) carefully before simplifying.
Remember \((1+n)^{3/2} = (\sqrt{1+n})^3\).

Answer 1

The Correct Option is D

Concept: The magnetic field on the axis of a circular current-carrying coil at a distance \(x\) from its center is given by: \[ B = \frac{B_0}{\left(1 + \dfrac{x^2}{R^2}\right)^{3/2}} \] where:

\(B_0\) is the magnetic field at the center of the coil,
\(R\) is the radius of the coil,
\(x\) is the axial distance from the center.
Step 1: Substitute the given values. \[ B_0 = 16~\mu\text{T}, \quad x = \sqrt{3}R \] \[ \frac{x^2}{R^2} = \frac{(\sqrt{3}R)^2}{R^2} = 3 \]
Step 2: Calculate the magnetic field at the given point. \[ B = \frac{16}{(1+3)^{3/2}} = \frac{16}{4^{3/2}} \] \[ 4^{3/2} = ( \sqrt{4} )^3 = 2^3 = 8 \] \[ B = \frac{16}{8} = 2~\mu\text{T} \]
Conclusion: \[ \boxed{B = 2~\mu\text{T}} \] Hence, the correct answer is (4).