Question

One of the two identical conducting wires of length L is bent in the form of a circular loop and the other one into a circular coil of N identical turns. If the same current is passed in both, the ratio of the magnetic field at the central of the loop (B_L) to that at the centre of the coil (B_C), i.e. \(\frac{{{{\rm{B}}_{\rm{L}}}}}{{{{\rm{B}}_{\rm{C}}}}}\) will be:
1. N
2. \(\frac{1}{{\rm{N}}}\)
3. N²
4. \(\frac{1}{{{{\rm{N}}^2}}}\)

Answer 1

Correct Answer - Option 4 : \(\frac{1}{{{{\rm{N}}^2}}}\)

Concept:

The wire of length L is made into circular loop.

Let the radius of the loop be R₁

The length of the wire = The circumference of the loop

L = 2πR₁

\(\Rightarrow {{\rm{R}}_1} = \frac{{\rm{L}}}{{2{\rm{\pi }}}}\) 

The magnetic field B_L at the centre of the loop is given by the formula:

\({{\rm{B}}_{\rm{L}}} = \frac{{{{\rm{\mu }}_0}{\rm{I}}}}{{2{{\rm{R}}_1}}}\) 

\(\Rightarrow {{\rm{B}}_{\rm{L}}} = \frac{{{{\rm{\mu }}_0}{\rm{I}}}}{{2\left( {\frac{{\rm{L}}}{{2{\rm{\pi }}}}} \right)}}\) 

\(\therefore {{\rm{B}}_{\rm{L}}} = \frac{{{{\rm{\mu }}_0}{\rm{I\pi }}}}{{\rm{L}}}\)     ----(1)

The wire of length L is made into circular coil with N turns. Let the radius of the coil be R₂

The length of the wire = The circumference of the coil

L = N(2πR₂)

\(\Rightarrow {{\rm{R}}_2} = \frac{{\rm{L}}}{{2{\rm{\pi N}}}}\) 

The magnetic field B_C at the centre of the coil is given by the formula:

\({{\rm{B}}_{\rm{C}}} = {\rm{N}}\left( {\frac{{{{\rm{\mu }}_0}{\rm{I}}}}{{2{{\rm{R}}_2}}}} \right)\) 

\(\Rightarrow {{\rm{B}}_{\rm{C}}} = {\rm{N}}\left( {\frac{{{{\rm{\mu }}_0}{\rm{I}}}}{{2\left( {\frac{{\rm{L}}}{{2{\rm{\pi N}}}}} \right)}}} \right)\) 

\(\therefore {{\rm{B}}_{\rm{C}}} = \frac{{{\rm{N}}{{\rm{\mu }}_0}{\rm{I}}\left( {{\rm{\pi N}}} \right)}}{{\rm{L}}}\) 

Calculation:

Now, from the question,

\(\Rightarrow \frac{{{{\rm{B}}_{\rm{L}}}}}{{{{\rm{B}}_{\rm{C}}}}} = \frac{{\frac{{{{\rm{\mu }}_0}{\rm{I\pi }}}}{{\rm{L}}}}}{{\frac{{{\rm{N}}{{\rm{\mu }}_0}{\rm{I}}\left( {{\rm{\pi N}}} \right)}}{{\rm{L}}}}}\) 

\(\Rightarrow \frac{{{{\rm{B}}_{\rm{L}}}}}{{{{\rm{B}}_{\rm{C}}}}} = \frac{{{{\rm{\mu }}_0}{\rm{I\pi }}}}{{\rm{L}}} \times \frac{{\rm{L}}}{{{\rm{N}}{{\rm{\mu }}_0}{\rm{I}}\left( {{\rm{\pi N}}} \right)}}\) 

\(\therefore \frac{{{{\rm{B}}_{\rm{L}}}}}{{{{\rm{B}}_{\rm{C}}}}} = \frac{1}{{{{\rm{N}}^2}}}\)