Question

Two concentric loops, each of radius equal to $2\pi \text{cm}$ are placed at right angles to each other. $3\text{A}$ and $4\text{A}$ are the currents flowing in each loop, respectively. The magnetic induction at the common center of the loops will be (in ${\text{Wb m}}^{-2}$)

• A. $5\times {10}^{-5}$
• B. $7\times {10}^{-5}$
• C. $12\times {10}^{-5}$
• D. ${10}^{-5}$
  

Answer 1

The correct option is A $5\times {10}^{-5}$


Magnetic field at the center of the loop is,

$B=\frac{{\mu }_{0}i}{2r}$

$\Rightarrow B_1=\frac{{\mu }_0{i}_1}{2r}$

$\Rightarrow B_2=\frac{{\mu }_0{i}_2}{2r}$

As we can see, at the common center, field due to loop $1$ and loop $2$ are right angle to each other i.e., $\theta ={90}^{\circ }$

$B_{net}=\sqrt{B_1^2+{B_2}^2}$

$=\sqrt{{\left(\frac{{\mu }_0{i}_1}{2r}\right)}^2+{\left(\frac{{\mu }_0{i}_2}{2r}\right)}^2}$

$=\frac{{\mu }_0}{2r}\sqrt{i_1^2+i_2^2}$

$=\frac{4\pi \times {10}^{-7}}{2\times 2\pi \times {10}^{-2}}\sqrt{3^2+4^2}$

$=5\times {10}^{-5}\text{T}\text{(or)}$

$=5\times {10}^{-5}{\text{Wb m}}^{-2}$

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