Question

Figure shows solenoid with 'n' turns per unit length and current 'i'. Magnetic field at P can be given as

• A. ${\mu }_{0}ni[\frac{1}{\sqrt{5}}-\frac{1}{2\sqrt{2}}]$
• B. $\frac{{\mu }_{0}ni}{\sqrt{2}}$
• C. ${\mu }_{0}ni[\frac{1}{\sqrt{2}}-\frac{1}{2}]$
• D. $\frac{{\mu }_{0}ni}{\sqrt{5}}$

Answer 1

The correct option is A ${\mu }_{0}ni[\frac{1}{\sqrt{5}}-\frac{1}{2\sqrt{2}}]$


Field at P due to element dx is
$d_B=\frac{{\mu }_0}{2}\frac{\left(di\right)R^2}{(R^2+x^2{)}^{\frac{3}{2}}}=\frac{{\mu }_0}{2}\frac{ndxi{R}^2}{(R^2+x^2{)}^{\frac{3}{2}}}$
$tan\theta =\frac{x}{R}$
$dx=Rse{c}^2\theta d\theta$
$B=\int dB=\frac{{\mu }_0}{2}ni{R}^2\int \frac{dx}{(R^2+x^2{)}^{\frac{3}{2}}}$
$B=\frac{{\mu }_0}{2}ni{R}^2{\int }_{{\theta }_1}^{{\theta }_2}\frac{Rse{c}^2\theta d\theta }{(R^2+R^{2}ta{n}^2\theta {)}^{\frac{3}{2}}}=\frac{{\mu }_0}{2}ni[sin{\theta }_2-sin{\theta }_1]$