question_answer

The magnetic field \[{{B}_{0}}\]due to current carrying circular loop of radius 12 cm at its centre is \[0.50\times 104\text{ }T.\]The magnetic field due to this loop at a point on the axis at a distance of 5 cm from the centre is:

A) \[3.9\times {{10}^{-5}}T\]            

B) \[9.3\times {{10}^{-5}}T\]

C) \[6.3\times {{10}^{-15}}T\]         

D) \[3.6\times {{10}^{-5}}T\]

Correct Answer: A

Solution :

Let there be a circular coil of radius a carrying a current i. Let P be a point on the axis of the coil at a distance \[x\] from the centre. The magnetic field at P is \[{{B}_{2}}=\frac{{{\mu }_{0}}Ni{{a}^{2}}}{2{{({{a}^{2}}+{{x}^{2}})}^{3/2}}}N{{A}^{-1}}{{m}^{-1}}\]                ?(i) Megnetic field at centre of coil, \[x=0\] \[\therefore \]      \[{{B}_{1}}=\frac{{{\mu }_{0}}Ni}{2a}N{{A}^{-1}}{{m}^{-1}}\]  ?(ii) Dividing Eq. (i) by (ii), we have \[\frac{{{B}_{2}}}{{{B}_{1}}}=\frac{{{a}^{3}}}{{{({{a}^{2}}+{{x}^{2}})}^{3/2}}}\] Putting the numerical values, we have \[\therefore \]     \[\frac{{{B}_{2}}}{{{B}_{1}}}=\frac{{{(0.12)}^{3}}}{{{[{{(0.12)}^{2}}+{{(0.05)}^{2}}]}^{3/2}}}\]             \[=\frac{{{(0.12)}^{3}}}{{{(0.0169)}^{3/2}}}\] \[\Rightarrow \]       \[B{{}_{2}}={{B}_{1}}\times {{\left( \frac{0.12}{0.13} \right)}^{3}}\] \[=0.50\times {{10}^{-4}}\times {{\left( \frac{0.12}{0.13} \right)}^{3}}\] \[=3.9\times {{10}^{-5}}\text{T}\]