Question

A current of $0.24\,A$ flows through a circular coil of 72 turns, the average diameter of the coil being 20 cm. What is the strength of the field produced at the centre of the coil?
A. $1.09 \times {10^{ - 4}}\,T$
B. $1.09 \times {10^{ - 3}}\,T$
C. $1.09 \times {10^{ - 2}}\,T$
D. $1.5 \times {10^{ - 4}}\,T$

Answer 1

Use Biot Savart’s law to determine the magnetic field at the center of the coil.
$B = \dfrac{{{\mu _0}nI}}{{2a}}$
Here, ${\mu _0}$ is the permeability of the vacuum, n is the number of turns, I is the current and a is the radius of the coil.

Complete step by step answer:
According to Biot Savart’s law, the magnetic field produced at the center of the coil of radius $a$, due to the current $I$ flowing through the coil is expressed as,
$B = \dfrac{{{\mu _0}nI}}{{2a}}$

Here, ${\mu _0}$ is the permeability of the vacuum, n is the number of turns of the coil, I is the current flowing through the coil and a is the radius of the coil.

The vacuum permeability is the constant and it has value $4\pi \times {10^{ - 7}}\,H/m$.

The diameter of the coil is twice the radius of the coil. Therefore, substitute d for 2a in the above equation.
$B = \dfrac{{{\mu _0}nI}}{d}$

Substitute $4\pi \times {10^{ - 7}}\,H/m$ for ${\mu _0}$, 72 for n, $0.24\,A$ for I, and 20 cm for d in the above equation.
$B = \dfrac{{\left( {4\pi \times {{10}^{ - 7}}\,H/m} \right)\left( {72} \right)\left( {0.24\,A} \right)}}{{\left( {20\,cm} \right)\left( {\dfrac{{{{10}^{ - 2}}\,m}}{{1\,cm}}} \right)}}$
$\Rightarrow B = \dfrac{{2.17 \times {{10}^{ - 5}}}}{{20 \times {{10}^{ - 2}}}}$
$\therefore B = 1.09 \times {10^{ - 4}}\,T$

So, the correct answer is “Option A”.

Note:
The modified form of the Biot Savart’s law, $B = \dfrac{{{\mu _0}nI}}{d}$ is used to calculate the field only at the centre of the coil. To calculate the magnetic field at other places, you need to use ideal Biot Savart’s law, $B = \dfrac{{{\mu _0}I}}{{4\pi }}\int {\dfrac{{d\vec l \times \vec r}}{{{a^2}}}}$.