Physics - Magnetism Question with Solution | TestHub

Magnetic field due to a long solenoid is given by $B={\mu }_{0}ni$ For solenoid with 200 turns per $\mathrm{cm}$ and $i$ current flows through it, $6.28\times 10^{-2}\mathrm{Wb}/{\mathrm{m}}^2={\mu }_0\times 200\times i$Now, for solenoid with 100 turns per cm with current $\frac{i}{3}$, $B={\mu }_0\times 100\times \left(\frac{i}{3}\right)$ Introducing value of $i$ from equation (1), $B={\mu }_0\times 100\times \left(\frac{1}{3}\times \frac{6.28\times 10^{-2}\mathrm{Wb}/{\mathrm{m}}^2}{{\mu }_0\times 200}\right)=1.46\times 10^{-2}\mathrm{Wb}/{\mathrm{m}}^2$ $B\approx 1.05\times 10^{-2}\mathrm{Wb}/{\mathrm{m}}^2$