Question

Find the magnitude of net magnetic field at point $\text{P}$ as shown in figure. If two loops of wire carry the same current of $10\text{mA}$, but the flow of current is in opposite directions. One loop has a radius of $R_1=50\text{cm}$ and the other loop has a radius of $R_2=100\text{cm}$. The distance of point $\text{P}$ from first loop is $0.25\text{m}$ and from second loop is $0.75\text{m}$.

[Take, $({0.25}^2+{0.5}^2{)}^{3/2}=0.175;(1+{0.75}^2{)}^{3/2}=1.95$]

• A. $2.8\times {10}^{-7}\text{T}$
• B. $5.77\times {10}^{-10}\text{T}$
• C. $6.91\times {10}^{-7}\text{T}$
• D. $8.75\times {10}^{-12}\text{T}$

Answer 1

The correct option is B $5.77\times {10}^{-10}\text{T}$

The net magnetic field at point $\text{P}$ will be obtained by super position of magnetic field produced due to each coil.

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Magnetic field due to a circular ring on the axis is given by

$B=\frac{{\mu }_{0}I{R}^2}{2(R^2+x^2{)}^{3/2}}$

Here,
$R_1=50\text{cm},x_1=0.25\text{m}$ and $R_2=100\text{cm},x_2=0.75\text{m}$

Thus, $(R_2^2+x_2^2{)}^{3/2}>(R_1^2+x_1^2{)}^{3/2}$

$\therefore B_1>B_2$

The direction of current in both coils is opposite, hence the produced magnetic field at $\text{P}$ will also be opposite.

$\therefore B_{net}=B_1-B_2$

$B_{net}=\frac{{\mu }_{0}I}{2}[\frac{R_1^2}{(R_1^2+x_1^2{)}^{3/2}}-\frac{R_2^2}{(R_2^2+x_2^2{)}^{3/2}}]$

$B_{net}=\frac{4\pi \times {10}^{-7}\times \left(0.001\right)}{2}[\frac{(0.5{)}^2}{({0.5}^2+{0.25}^2{)}^{3/2}}-\frac{(1{)}^2}{(1^2+{0.75}^2{)}^{3/2}}]$

$B_{net}=0.002\pi \times {10}^{-7}[1.431-0512]$

$\therefore B_{net}=5.77\times {10}^{-10}\text{T}$

Hence, option $\left(b\right)$ is correct.