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Two similar coils which are separated by $2 \mathrm{~m}$ having radius of $1 \mathrm{~m}$ and number of turns 80 have a common axis. Calculate the strength of magnetic field in microtesla at a point midway between them on their common axis when a current is $0.2$
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Verified Answer
The correct answer is:
$1.6$
The given situation is shown below,
Net magnetic field, $\mathbf{B}_{\text {net }}=\mathbf{B}_1+\mathbf{B}_2$
Since both the magnetic fields are in same directions.
So, the magnitude of net magnetic field can be calculated by algebric addition of magnitudes of $\mathbf{B}_1$ and $\mathbf{B}_2$.
Hence, $B_{\text {net }}=B_1+B_2$
Here,
$\begin{aligned}
B_1 & =\frac{\mu_0 N_1 I_1 R_1^2}{2\left(R_1^2+x^2\right)^{3 / 2}}=\frac{4 \pi \times 10^{-7} \times 80 \times 0.2 \times(1)^2}{2\left(1^2+1^2\right)^{3 / 2}} \\
& =\frac{32 \pi \times 10^{-7}}{\left(2^{3 / 2}\right.}=\frac{32 \pi \times 10^{-7}}{2 \sqrt{2}}=8 \sqrt{2} \times 10^{-7} \mathrm{~T}
\end{aligned}
$
and
$
\begin{aligned}
B_2 & =\frac{\mu_0 N_2 \cdot I_2 R_2^2}{2\left(R_2^2+x^2\right)^{3 / 2}}=\frac{4 \pi \times 10^{-7} \times 80 \times 0.2 \times(1)^2}{2\left(1^2+1^2\right)^{3 / 2}} \\
& =8 \sqrt{2} \times 10^{-7} \mathrm{~T}
\end{aligned}
$
Therefore,
$
\begin{aligned}
B_{\text {net }} & =8 \sqrt{2} \times 10^{-7}+8 \sqrt{2} \times 10^{-7} \\
& =16 \sqrt{2} \times 10^{-7} \mathrm{~T}
\end{aligned}$
Net magnetic field, $\mathbf{B}_{\text {net }}=\mathbf{B}_1+\mathbf{B}_2$
Since both the magnetic fields are in same directions.
So, the magnitude of net magnetic field can be calculated by algebric addition of magnitudes of $\mathbf{B}_1$ and $\mathbf{B}_2$.
Hence, $B_{\text {net }}=B_1+B_2$
Here,
$\begin{aligned}
B_1 & =\frac{\mu_0 N_1 I_1 R_1^2}{2\left(R_1^2+x^2\right)^{3 / 2}}=\frac{4 \pi \times 10^{-7} \times 80 \times 0.2 \times(1)^2}{2\left(1^2+1^2\right)^{3 / 2}} \\
& =\frac{32 \pi \times 10^{-7}}{\left(2^{3 / 2}\right.}=\frac{32 \pi \times 10^{-7}}{2 \sqrt{2}}=8 \sqrt{2} \times 10^{-7} \mathrm{~T}
\end{aligned}
$
and
$
\begin{aligned}
B_2 & =\frac{\mu_0 N_2 \cdot I_2 R_2^2}{2\left(R_2^2+x^2\right)^{3 / 2}}=\frac{4 \pi \times 10^{-7} \times 80 \times 0.2 \times(1)^2}{2\left(1^2+1^2\right)^{3 / 2}} \\
& =8 \sqrt{2} \times 10^{-7} \mathrm{~T}
\end{aligned}
$
Therefore,
$
\begin{aligned}
B_{\text {net }} & =8 \sqrt{2} \times 10^{-7}+8 \sqrt{2} \times 10^{-7} \\
& =16 \sqrt{2} \times 10^{-7} \mathrm{~T}
\end{aligned}$
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