The radius of a circular current carrying coil is R. At what distance from the centre of the coil on its axis, the intensity of magnetic field will be (1/2 √2) times that at the centre?

Question

The radius of a circular current carrying coil is R. At what distance from the centre of the coil on its axis, the intensity of magnetic field will be (1/2 √2) times that at the centre? (A) 2 R (B) (3 R/2) (C) R (D) (R/2)

Tardigrade Admin · Accepted Answer

(μ0 i R2/2(R2+x2)3 / 2)=(1/2 √2) (μ0 i/2 R) 2 √2 R3=(R2+x2)3 / 2 (2 √2)2 / 3 R2=R2+x2 x2=2 R2-R2=R2 ⇒ x=R

Q. The radius of a circular current carrying coil is $R$ . At what distance from the centre of the coil on its axis, the intensity of magnetic field will be $\frac{1}{2 2}$ times that at the centre?

$2 R$

$\frac{3 R}{2}$

$R$

$\frac{R}{2}$

$\frac{μ _{0} i R ^{2}}{2 ( R ^{2} + x ^{2} ) ^{3/2}} = \frac{1}{2 2} \frac{μ _{0} i}{2 R}$
$22 R^{3} = (R^{2} + x^{2})^{3/2}$
$(22)^{2/3} R^{2} = R^{2} + x^{2}$
$x^{2} = 2 R^{2} - R^{2} = R^{2}$
$\Rightarrow x = R$

Q. The radius of a circular current carrying coil is R. At what distance from the centre of the coil on its axis, the intensity of magnetic field will be 22​1​ times that at the centre?

2R

23R​

R

2R​

2(R2+x2)3/2μ0​iR2​=22​1​2Rμ0​i​ 22​R3=(R2+x2)3/2 (22​)2/3R2=R2+x2 x2=2R2−R2=R2 ⇒x=R

Q. The radius of a circular current carrying coil is $R$ . At what distance from the centre of the coil on its axis, the intensity of magnetic field will be $\frac{1}{2 2}$ times that at the centre?

$2 R$

$\frac{3 R}{2}$

$R$

$\frac{R}{2}$

$\frac{μ _{0} i R ^{2}}{2 ( R ^{2} + x ^{2} ) ^{3/2}} = \frac{1}{2 2} \frac{μ _{0} i}{2 R}$
$22 R^{3} = (R^{2} + x^{2})^{3/2}$
$(22)^{2/3} R^{2} = R^{2} + x^{2}$
$x^{2} = 2 R^{2} - R^{2} = R^{2}$
$\Rightarrow x = R$