Question

A circular coil carrying a current I has radius R and number of turns N . If all the three, i.e., the current I, radius R and number of turns N are doubled, then magnetic field at its centre becomes
(A) Double
(B) Half
(C) Four times
(D) One fourth

Answer 1

Hint: When the current I is flowing in the circular coil of radius R and number of turns N then first we use Biot savart law to derive the relation between magnetic field, radius, current and number of turns i.e., $B = \dfrac{{{\mu _0}NI{R^2}}}{{2{{({R^2} + {x^2})}^{3/2}}}}$ and use above expression for centre point.

Complete step by step answer:
We know that at the axis of a current carrying coil at distance x from the centre, magnetic field B is given by
$B = \dfrac{{{\mu _0}NI{R^2}}}{{2{{({R^2} + {x^2})}^{3/2}}}}$
At centre $x = 0$
So, ${B_C} = \dfrac{{{\mu _0}NI{R^2}}}{{2{{({R^2})}^{3/2}}}}$
${B_C} = \dfrac{{{\mu _0}NI{R^2}}}{{2{R^3}}}$
$\implies {B_C} = \dfrac{{{\mu _0}IN}}{{2R}}$ …..(1)
If I, R & N are doubled then magnetic field at centre will be
${B_C}’$ = $\dfrac{{{\mu _0}(2I)(2N)}}{{2(2R)}}$
$\implies {B_C}’$ = $2\left( {\dfrac{{{\mu _0}NI}}{{2R}}} \right)$
From equation (1)
${B_C}’$ = $2{B_C}$
Hence, the magnetic field at the centre will also become doubled.

So, option A is the correct answer Double.

Note:
If any point which is situated at any distance from centre on axial line, then we can also calculate the magnetic field by above formula which is
$B = \dfrac{{{\mu _0}NI{R^2}}}{{2{{({R^2} + {x^2})}^{3/2}}}}$.