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

Hint: 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, \[B = \dfrac{{{\mu _0}I}}{{4\pi }}\int {\dfrac{{d\vec l \times \vec r}}{{{a^2}}}} \].