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 \[\therefore \] \[\frac{{{B}_{2}}}{{{B}_{1}}}=\frac{{{(0.12)}^{3}}}{{{[{{(0.12)}^{2}}+{{(0.05)}^{2}}]}^{3/2}}}\] \[=\frac{{{(0.12)}^{3}}}{{{(0.0169)}^{3/2}}}\] \[\Rightarrow \] \[B{{}_{2}}={{B}_{1}}\times {{\left( \frac{0.12}{0.13} \right)}^{3}}\] \[=0.50\times {{10}^{-4}}\times {{\left( \frac{0.12}{0.13} \right)}^{3}}\] \[=3.9\times {{10}^{-5}}\text{T}\]You need to login to perform this action.
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