Question

A circular loop of radius 4.0 cm is placed in a horizontal plane and carries an electric current of 5.0 A in the clockwise direction as seen from above. Find the magnetic field (a) at a point 3.0 cm above the centre of the loop (b) at a point 3.0 cm below the centre of the loop.

Answer 1

Explanation:

Given data in the question  

Current Magnitude I = 5.0 A

Loop Radius, r = 4.0 cm

(a) To find: The magnetic field at a point 3.0 cm above the centre of the loop

As we know that ,

$B=\frac{\mu_{0} i r^{2}}{2\left(x^{2}+r^{2}\right)^{\frac{3}{2}}}$

$=\frac{5 \times 4 \pi \times 10^{-7} \times 16 \times 10^{-4}}{2\left[(9 \times 16) \times 10^{-4}\right]^{\frac{3}{2}}}$

$=\frac{4 \pi \times 80 \times 10^{-11}}{2\left[(144) \times 10^{-4}\right]^{\frac{3}{2}}}$

$=\frac{4 \pi \times 80 \times 10^{-11}}{2 \times 125 \times 10^{-6}}$

$=4.019 \times 10^{-5} T$    (Above the middle of the loop )

(b) To find: The magnetic field strength B at point O given at a point 3.0 cm below the centre of the loop.

$B=\frac{\mu_{0} i r^{2}}{2\left(x^{2}+r^{2}\right)^{\frac{3}{2}}}$

$=\frac{5 \times 4 \pi \times 10^{-7} \times 16 \times 10^{-4}}{2\left[(9 \times 16) \times 10^{-4}\right]^{\frac{3}{2}}}$

$=\frac{4 \pi \times 80 \times 10^{-11}}{2\left[(144) \times 10^{-4}\right]^{\frac{3}{2}}}$

$=\frac{4 \pi \times 80 \times 10^{-11}}{2 \times 125 \times 10^{-6}}$

$=4.019 \times 10^{-5} T$   (Below the middle of the loop )