Question

A winding wire which is used to frame a solenoid can bear a maximum 10Acurrent. If length of solenoid is 80 cm and its cross-sectional radius is 3 cm, then required length of winding wire is (Magnetic field due to solenoid = 0.2 T)

• 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 $

Answer 1

Answer: (B)

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 $