At what distance on the axis, from the centre of a circular current carrying coil of radius r, the magnetic field becomes 1 / 8 th of the magnetic field at centre?

Question

At what distance on the axis, from the centre of a circular current carrying coil of radius r, the magnetic field becomes 1 / 8 th of the magnetic field at centre? (A) √2 r (B) 23 / 2 r (C) √3 r (D) 3 √2 r

Tardigrade Admin · Accepted Answer

(1/8) (μ0 i/2 r)=(μ0 r2/2(r2+x2)3 / 2) ⇒ (r2+x2)3 / 2=8 r3 ⇒ (r2+x2)=(8 r3)2 / 3 ⇒ r2+x2=4 r2 ⇒ 3 r2=x2 ⇒ x=r √3

Q. At what distance on the axis, from the centre of a circular current carrying coil of radius $r$ , the magnetic field becomes $1/8$ th of the magnetic field at centre?

$2 r$

$2^{3/2} r$

$3 r$

$32 r$

$\frac{1}{8} \frac{μ _{0} i}{2 r} = \frac{μ _{0} r ^{2}}{2 ( r ^{2} + x ^{2} ) ^{3/2}}$
$\Rightarrow (r^{2} + x^{2})^{3/2} = 8 r^{3}$
$\Rightarrow (r^{2} + x^{2}) = (8 r^{3})^{2/3}$
$\Rightarrow r^{2} + x^{2} = 4 r^{2}$
$\Rightarrow 3 r^{2} = x^{2}$
$\Rightarrow x = r 3$

Q. At what distance on the axis, from the centre of a circular current carrying coil of radius r, the magnetic field becomes 1/8 th of the magnetic field at centre?

2​r

23/2r

3​r

32​r