A) \[1.2\times {{10}^{2}}m\]
B) \[4.8\times {{10}^{2}}m\]
C) \[2.4\times {{10}^{3}}m\]
D) \[6\times {{10}^{3}}m\]
Correct Answer: C
Solution :
\[B=\frac{{{\mu }_{0}}Ni}{l}\]; where N = total number of turns, \[l=\] length of the solenoid \[\Rightarrow \] \[0.2=\frac{4\pi \times {{10}^{-7}}\times N\times 10}{0.8}\] \[\Rightarrow \] \[N=\frac{4\times {{10}^{4}}}{\pi }\] Since N turns are made from the winding wire so length of the wire \[(L)=2\pi r\times N\][\[2\pi r\]= length of each turns] \[\Rightarrow \] \[L=2\pi \times 3\times {{10}^{-2}}\times \frac{4\times {{10}^{4}}}{\pi }\] \[=2.4\times {{10}^{3}}m\]
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