A) \[1.2\times {{10}^{2}}\,m\]
B) \[2.4\times {{10}^{3}}\,m\]
C) \[4.8\times {{10}^{2}}\,m\]
D) \[6\times {{10}^{3}}\,m\]
Correct Answer: B
Solution :
The magnetic field due to a solenoid. \[B=\frac{{{\mu }_{0}}Ni}{l}\]where \[N=\]number of turns and \[l=\]length of solenoid. \[0.2=\frac{4\pi \times {{10}^{-7}}\times N\times 10}{0.8}\] \[N=\frac{0.8\times 0.2}{4\times \pi \times {{10}^{-6}}}\] \[=\frac{4\times {{10}^{4}}}{\pi }\] Total length of the wire\[(L)=2\pi r\times N\] \[=2\pi \times 3\times {{10}^{-2}}\times \frac{4\times {{10}^{4}}}{\pi }\] \[=2.4\times {{10}^{3}}m\]
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