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CMC Medical CMC-Medical Ludhiana Solved Paper-2015

  • question_answer
    An \[{{N}_{1}}\text{-}\]turns coil having diameter D carries a current\[I\]. An another coil is made of same material having number of turns \[{{N}_{2}},\] diameter 2D and carrying the same current\[I\]. If the length of wire in both coils are same then ratio of their magnetic moments\[\left( \frac{{{M}_{1}}}{{{M}_{2}}} \right)\]be

    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, \[M=NIA\] \[\therefore \]  \[\frac{{{M}_{1}}}{{{M}_{2}}}=\frac{{{N}_{1}}I{{A}_{1}}}{{{N}_{2}}I{{A}_{2}}}=\frac{{{N}_{1}}\pi D_{1}^{2}}{{{N}_{2}}\pi D_{2}^{2}}\]                        \[=\frac{{{N}_{1}}\pi D_{1}^{2}}{{{N}_{2}}\pi 4D_{1}^{2}}\]                   \[(\because \,{{D}_{2}}=2{{D}_{1}})\]        \[=\left( \frac{{{N}_{1}}}{{{N}_{2}}} \right)\times \frac{1}{4}=\frac{2}{1}\times \frac{1}{4}=\frac{1}{2}\] \[\Rightarrow \]               \[{{M}_{1}}:{{M}_{2}}=1:2\]


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