Question

The field normal to the plane of a wire of n turns and radius r which carries a current I is measured of the coil at a small distance h from the centre of the coil. This is smaller than the field at the centre by the fraction –

• A. $\frac { 3 } { 2 }$ $\frac{h^{2}}{r^{2}}$
• B. $\frac { 2 } { 3 }$ $\frac{h^{2}}{r^{2}}$
• C. $\frac { 3 } { 2 }$ $\frac{r^{2}}{h^{2}}$
• D. $\frac { 2 } { 3 }$ $\frac{r^{2}}{h^{2}}$

Answer 1

Answer: (A)

\frac { 3 } { 2 }$$\frac{h^{2}}{r^{2}}

Answer 2

$\frac{B_{axis}}{B_{centre}} = \frac{r^{3}}{(r^{2} + x^{2})^{3/2}}$

B_axis = B_centre = $\frac{r^{3}}{r^{3}\left( 1 + \frac{x^{2}}{r^{2}} \right)^{3/2}}$

= B_c$\left( 1 + \frac{x^{2}}{r^{2}} \right)^{- 3/2}$

x = h B_axis = $\left( 1 - \frac{3h^{2}}{2r^{2}} \right)$B_c Ž DB = B_c – B_axis = $\frac{3}{2}\frac{h^{2}}{r^{2}}$