Question

two circular loop of radii 10 m and 40 cm carries current of 1 A and 2 A respetively.find the ratio of magnitude of magnetic field at their centre

Answer 1

we know,
magnetic field at the centre of circular coil of radius r , number of turns N, and current I passing through the coil/loop is
$\qquad\bf{B=\frac{\mu_0NI}{2r}}$

so, for getting ratio of magnetic field of two given loop , use formula
$\qquad\bf{\frac{B_1}{B_2}=\frac{I_1}{I_2}\frac{r_2}{r_1}}$

here, $I_1=1A,I_2=2A$
$r_1=10cm,r_2=40cm$

so, $\frac{B_1}{B_1}=\frac{1}{2}\frac{40}{10}$

$\frac{B_1}{B_2}=\frac{2}{1}$

Answer 2

Typing mistake in the question :

It will be 10 cm instead of 10 m .

Given information :

Radius of the first circular loop ( r₁ ) = 10 cm .

Radius of the second circular loop ( r₂ ) = 40 cm .

Current in the first loop ( I₁ ) = 1 A .

Current in the second loop ( I₂ ) = 2 A .

We will use the Biot Savart's Law in this numerical .

Let the current be I , number of turns be N , magnetic field strength be B , r be the radius of the coil , then we have :

$\mathsf{B=\dfrac{\mu_0NI}{2r}}$

Here μ₀ is a constant which has a value of  4 π × 10⁻⁷ .

Using the above formula :

$\mathsf{\dfrac{B_1}{B_2}=\dfrac{\dfrac{\mu_0NI_1}{2r_1}}{\dfrac{\mu_0NI_2}{2r_2}}}\\\\\implies \mathsf{\dfrac{B_1}{B_2}=\dfrac{I_1r_2}{I_2r_1}}\\\\\implies \mathsf{\dfrac{B1}{B_2}=\dfrac{1A\times 40cm}{2A\times 10 cm}}\\\\\implies \mathsf{\dfrac{B_1}{B_2}=\dfrac{40}{20}}\\\\\implies \mathsf{\dfrac{B_1}{B_2}=\dfrac{2}{1}}\\\\\implies \boxed{\mathtt{B_1:B_2=2:1}}$

The required ratio is 2 : 1 .