Question

Magnetc field due to a current carrying circular loop of radius $12\text{cm}$ at its centre is $0.5\times {10}^{-4}\text{T}$. Find the magnetic field due to this loop at a point on the axis at a distance $5.0\text{cm}$ from centre.

• A. $3.9\times {10}^{-5}\text{T}$
• B. $7.8\times {10}^{-5}\text{T}$
• C. $11.7\times {10}^{-5}\text{T}$
• D. None of these

Answer 1

The correct option is A $3.9\times {10}^{-5}\text{T}$

Magnetic field at the centre of the loop is given by,

$B_1=\frac{{\mu }_{0}I}{2R}=0.5\times {10}^{-4}\text{T}.......\left(1\right)$

For an axial point,

$B_2=\frac{{\mu }_{0}I{R}^2}{2{(R^2+x^2)}^{3/2}}........\left(2\right)$

From equation $\left(1\right)\&\left(2\right)$, we get

$\frac{B_1}{B_2}=\frac{(\frac{{\mu }_{0}I}{2R})}{(\frac{{\mu }_{0}I{R}^2}{2{(R^2+x^2)}^{3/2}})}$

$\frac{B_1}{B_2}=\frac{{(R^2+x^2)}^{3/2}}{R^3}$

$B_2=B_1\left[\frac{R^3}{{(R^2+x^2)}^{3/2}}\right]$

Putting the given data, we get

$B_2=0.5\times {10}^{-4}\left[\frac{(12{)}^3}{({12}^2+5^2{)}^{\frac{3}{2}}}\right]$

$B_2=3.9\times {10}^{-5}\text{T}$

Why this Question ? Key point - The magnetic field due to a circular current carrying loop at an axial point is given by, $B_{axial}=\frac{{\mu }_{0}I{R}^2}{2{(R^2+x^2)}^{3/2}}$