Question

Two concentric coils each of radius equal to $2\pi \text{cm}$ are placed in such a manner that their planes are at right angles to each other. Current of $3\text{A}$ and $4\text{A}$ respectively are flowing through the coil. What will be the magnetic induction at the center of the loop ?

• A. $5\times {10}^{-5}\text{T}$
• B. $7\times {10}^{-5}\text{T}$
• C. $1\times {10}^{-5}\text{T}$
• D. None of these

Answer 1

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


The direction of the magnetic fields due to both the loops is determined by using right-hand thumb rule.

Magnetic field due to a loop-$\text{B}$ carrying current $4\text{A}$ is,

$B_1=\frac{{\mu }_{0}I}{2R}$

$B_1=\frac{4\pi \times {10}^{-7}\times 4}{2\times 2\pi \times {10}^{-2}}$

$\therefore B_1=4\times {10}^{-5}\text{T}$ along $y-$axis

So, magnetic field due to loop-$\text{A}$ carrying current $3\text{A}$ is,

$B_2=\frac{{\mu }_{0}I}{2R}$

$B_2=\frac{4\pi \times {10}^{-7}\times 3}{2\times 2\pi \times {10}^{-2}}$

$\therefore B_2=3\times {10}^{-5}\text{T}$ along $x-$axis

So the net magnetic field is given by,

$\overrightarrow{B_{\text{net}}}=\overrightarrow{B_1}+\overrightarrow{B_2}=B_1\hat{j}+B_2\hat{i}$

$\therefore \left|\begin{array}{c}\overrightarrow{B_{net}}\end{array}\right|=\sqrt{{\left|\begin{array}{c}{B}_1\end{array}\right|}^2+{\left|\begin{array}{c}{B}_2\end{array}\right|}^2}$

$B_{net}=\sqrt{(4\times {10}^{-5}{)}^2+(3\times {10}^{-5}{)}^2}$

$\therefore B_{net}=5\times {10}^{-5}\text{T}$

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