A) 1 : 2
B) 2 : 1
C) 4 : 1
D) 1 : 4
Correct Answer: A
Solution :
The lengths of wire are same in both coil. So, \[{{N}_{1}}\times \pi {{D}_{1}}={{N}_{2}}\times \pi {{D}_{2}}\] \[\Rightarrow \] \[{{N}_{1}}\times \pi {{D}_{1}}={{N}_{2}}\times 2\pi {{D}_{1}}\] \[(\because {{D}_{2}}=2{{D}_{1}})\] \[\frac{{{N}_{1}}}{{{N}_{2}}}=\frac{2}{1}=2:1\] Now magnetic moment, \[M=NIA\] \[\therefore \] \[\frac{{{M}_{1}}}{{{M}_{2}}}=\frac{{{N}_{1}}I{{A}_{1}}}{{{N}_{2}}I{{A}_{2}}}=\frac{{{N}_{1}}\pi D_{1}^{2}}{{{N}_{2}}\pi D_{2}^{2}}\] \[=\frac{{{N}_{1}}\pi D_{1}^{2}}{{{N}_{2}}\pi 4D_{1}^{2}}\] \[(\because \,{{D}_{2}}=2{{D}_{1}})\] \[=\left( \frac{{{N}_{1}}}{{{N}_{2}}} \right)\times \frac{1}{4}=\frac{2}{1}\times \frac{1}{4}=\frac{1}{2}\] \[\Rightarrow \] \[{{M}_{1}}:{{M}_{2}}=1:2\]You need to login to perform this action.
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