Question

Two circular loops of same radii, $R=1\text{m}$ carrying same current, $I=1\text{A}$ as shown in the figure. The effective distance between the centres of the loop is, $x=\sqrt{3}\text{m}$. Find the net magnetic field at point $\text{P}$.

• A. $6.33\times {10}^{-7}\text{T}$
• B. $63.33\times {10}^{-7}\text{T}$
• C. $633.3\times {10}^{-7}\text{T}$
• D. $0.633\times {10}^{-7}\text{T}$

Answer 1

The correct option is A $6.33\times {10}^{-7}\text{T}$

Given:

$R_1=R_2=1\text{m},I_1=I_2=1\text{A}$ and
$x=\sqrt{3}\text{m}$

The resultant magnetic field at point $\text{P}$ is due to both loop $\left(1\right)$ and loop $\left(2\right)$.

Magnetic field due to loop $1$,

$B_1=\frac{{\mu }_o{I}_1{R}_1^2}{2(R_1^2+x^2{)}^{3/2}}$

$\Rightarrow B_1=\frac{4\pi \times {10}^{-7}\times 1\times 1^2}{2[1^2+(\sqrt{3}{)}^2{]}^{3/2}}$

$\Rightarrow B_1=7.85\times {10}^{-8}\text{T}\left[\text{Toward +x axis}\right]$

Magnetic field due to loop $2$,

$B_2=\frac{{\mu }_o{I}_2}{2R_2}=\frac{4\pi \times {10}^{-7}\times 1}{2\times 1}$

$\Rightarrow B_2=6.28\times {10}^{-7}\text{T}\left[\text{Toward +y axis}\right]$

Now, net magnetic field,

$B=\sqrt{B_1^2+B_2^2}=\sqrt{(7.85\times {10}^{-8}{)}^2+(6.28\times {10}^{-7}{)}^2}$

$\Rightarrow B=6.33\times {10}^{-7}\text{T}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.