A) 10
B) \[10\sqrt{10}\]
C) 20\[\sqrt{5}\]
D) 20\[\sqrt{10}\]
Correct Answer: B
Solution :
The magnetic field at a point on the axis of circular coil is \[B=\frac{{{\mu }_{0}}ni\,{{R}^{2}}}{2{{({{R}^{2}}+{{x}^{2}})}^{3/2}}}\]where, R is radius of coil, \[x=\]distance of any point on the axis In the first case: R = 22R and \[x=0\]the magnetic field at the centre is \[{{B}_{1}}=\frac{{{\mu }_{0}}ni4{{R}^{2}}}{2{{(4{{R}^{2}}+{{0}^{2}})}^{3/2}}}\] \[\Rightarrow \] \[{{B}_{1}}=\frac{{{\mu }_{0}}\,ni\,4{{R}^{2}}}{2{{(4{{R}^{2}})}^{3/2}}}\] ?(i) In the second case: when \[x=6R\]and \[R=2R\] \[{{B}_{2}}=\frac{{{\mu }_{0}}ni\times 4{{R}^{2}}}{2{{(4{{R}^{2}}+36{{R}^{2}})}^{3/2}}}\] \[=\frac{{{\mu }_{0}}\,ni\times 4{{R}^{2}}}{2{{(40{{R}^{2}})}^{3/2}}}\] ?.(ii) The ratio of magnetic field \[{{B}_{1}}\]and \[{{B}_{2}}\]is \[\frac{{{B}_{1}}}{{{B}_{2}}}=\frac{{{\mu }_{0}}ni4{{R}^{2}}}{2\times {{4}^{3/2}}\times {{R}^{3}}}\times \frac{2\times {{40}^{3/2}}{{R}^{3}}}{{{\mu }_{0}}\,ni\,4{{R}^{2}}}\] \[={{\left( \frac{40}{4} \right)}^{3/2}}={{(10)}^{3/2}}\] \[=\sqrt{10}\times 10=10\sqrt{10}\]
You need to login to perform this action.
You will be redirected in
3 sec
