Question 130: Magnetism and Matter

Question

A coil of n number of turns wound tightly in the form of a spiral with inner and outer radii r1 and r2 respectively. When a current of strength I is passed through the coil, the magnetic field at its centre will be :

Accepted Answer

In the width of $${r_2} - {r_1}$$ total n turns presents.
$$	herefore$$ In 1 unit width $${n \over {{r_2} - {r_1}}}$$ turns presents.
$$	herefore$$ In the width of dr number of turns,
$$n' = {n \over {{r_2} - {r_1}}} 	imes dr$$
Magnetic field (dB) due to element of dr length is
$$dB = {{{\mu _0} 	imes I 	imes n'} \over {2r}}$$
$$ = {{{\mu _0}I} \over {2r}} 	imes {n \over {({r_2} - {r_1})}}$$
$$	herefore$$ Total magnetic field due to entire coil is,
$$\int {dB = \int_{{r_1}}^{{r_2}} {{{{\mu _0}I} \over {2r}} 	imes {n \over {({r_2} - {r_1})}}dr} } $$
$$ \Rightarrow B = {{{\mu _0}In} \over {2({r_2} - {r_1})}}\int_{{r_1}}^{{r_2}} {{{dr} \over r}} $$
$$ = {{{\mu _0}In} \over {2({r_2} - {r_1})}} 	imes \left[ {\log _e^r} ight]_{{r_2}}^{{r_1}}$$
$$ = {{{\mu _0}In} \over {2({r_2} - {r_1})}} 	imes \left( {\log _e^{{r_2}} - \log _e^{{r_1}}} ight)$$
$$ = {{{\mu _0}In} \over {2({r_2} - {r_1})}} 	imes {\log _e}{{{r_2}} \over {{r_1}}}$$

Answer

$${{{\mu _0}nI} \over {2({r_2} - {r_1})}}$$

Answer

$${{{\mu _0}nI} \over {{r_2}}}$$

Answer

$${{{\mu _0}nI} \over {{r_2} - {r_1}}}{\log _e}{{{r_1}} \over {{r_2}}}$$

Answer

$${{{\mu _0}nI} \over {2({r_2} - {r_1})}}{\log _e}{{{r_2}} \over {{r_1}}}$$

Magnetism and Matter

A coil of n number of turns wound tightly in the form of a spiral with inner and outer radii r1 and r2 respectively. When a current of strength I is passed through the coil, the magnetic field at its centre will be :

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