question_answer

A circular coil of radius R carries an electric current. The magnetic Held due to the coil at a point on the axis of the coil located at a distance  from the centre of the coil, such that \[r>>R,\] varies as :

A) 1/r       

B)                       \[1/{{r}^{3/2}}\]                              

C) \[1/{{r}^{2}}\]                   

D)       \[1/{{r}^{3}}\]

Correct Answer: D

Solution :

                Key Idea: For a circular coil, the component of the field B perpendicular to the axis at P cancel each other while along the axis add up. The resultant magnetic field at point P will be due to the components along the axis. Hence, \[B=\int{dB\,\sin \beta =\frac{{{\mu }_{0}}}{\mu \pi }}\int{\frac{i\,dl\,\sin \,\theta }{{{r}^{2}}}}\sin \beta \]s and as here angle \[\theta \] between the element \[d\overrightarrow{\mathbf{l}}\]and \[\overrightarrow{r}\] is \[\frac{\pi }{2}\] everywhere and r is same for all  elements while \[\sin \beta =\frac{R}{r},\] so Hence, we have \[B=\frac{{{\mu }_{0}}}{4\pi }\frac{2\pi i{{R}^{2}}}{{{x}^{3}}}\] where      \[x={{({{R}^{2}}+{{r}^{2}})}^{1/2}}\] \[B=\frac{{{\mu }_{0}}}{4\pi }\frac{2\pi i{{R}^{2}}}{{{({{R}^{2}}+{{r}^{2}})}^{3/2}}}\] Given, r> >R then we have, neglecting R, \[B=\frac{{{\mu }_{0}}}{4\pi }\frac{2\pi i{{R}^{2}}}{{{r}^{3}}}\] Also area \[=\pi {{R}^{2}}\] \[\therefore \]    \[B=\frac{{{\mu }_{0}}}{2\pi }\frac{Ai}{{{r}^{3}}}\] \[\Rightarrow \]   \[B\propto \frac{1}{{{r}^{3}}}\]