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 $(L)=2\pi r\times N$ $=2\pi \times 3\times {10}^{-2}\times \frac{4\times {10}^4}{\pi }$ $=2.4\times {10}^{3}m$