In Figure 29-46 two concentric circular loops of wire carrying current in the same direction lie in the same plane. Loop 1 has radius$\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{cm}$ and carries $\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{mA}$. Loop 2 has radius$\mathbf{}\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{cm}$and carries $\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{mA}$.Loop 2 is to be rotated about a diameter while the net magnetic field $\overrightarrow{\mathbf{B}}$set up by the two loops at their common center is measured. Through what angle must loop 2 be rotated so that the magnitude of that net field is $\mathbf{100}\mathbf{}\mathbf{nT}$?

1. Radius of loop 1$r_1=1.50\mathrm{cm}=1.50\times {10}^{-2}\mathrm{m}$.
2. Current in the loop 1$i_1=4.00\text{mA}=4.00\times {10}^{-3}\text{A}$.
3. Radius of loop 2$r_2=2.50\text{cm}=2.50\times {10}^{-2}\text{m}$.
4. Current in the loop 2$i_2=6.00\text{mA}=6.00\times {10}^{-3}\text{A}$.
5. Net magnetic field$B=100\text{nT}$.

Use the formula of the magnetic field at the center of the coil due to two wires. From this find the angle through which loop 2 must be rotated so that the magnitude of the net filed is $\mathbf{100}\mathbf{}\mathbf{nT}$.

Formula:

For the magnetic field at the center of loop 1.

$B_1=\frac{{\mu }_0{i}_1}{2r_1}$

$B_1=\frac{\left(1.26\times {10}^{-6}T.mA\right)\left(4.00\times {10}^{-3}A\right)}{2\left(1.50\times {10}^{-2}m\right)}$

$B_1=1.68\times {10}^{-7}T=168nT$

Now, for the magnetic field at the center of loop 2.

$B_2=\frac{{\mu }_0{i}_2}{2r_2}$

Substitute the values and solve as:

$B_2=\frac{\left(1.26\times {10}^{-6}\text{T}·\text{mA}\right)\left(6.00\times {10}^{-3}\text{A}\right)}{2\left(2.50\times {10}^{-2}\text{m}\right)}$

$$\begin{gathered} B_2=1.512\times {10}^{-7}\text{T} \\ {B}_2=151.2\text{nT} \end{gathered}$$

Now, we use the cosine rule of vector addition

$B^2=B_1^2+B_2^2+2B_1{B}_2\cos \theta$

${\left(100\text{nT}\right)}^2={\left(168\text{nT}\right)}^2+{\left(151.2\right)}^2+2\left(168\text{nT}\right)\left(151.2\right)\cos \theta$

$\left(10000\mathrm{nT}-51085.44\mathrm{nT}\right)=\left(50803.2\mathrm{nT}\right)\cos \theta$

$\left(50803.2\text{nT}\right)\cos \theta =\left(-41085.44\text{nT}\right)$

Solve further as:

$\cos \theta =\frac{-41085.44nT}{\left(50803.2nT\right)}$

$\cos \theta =-0.81$

$\theta ={\cos }^{-1}\left(-0.81\right)$

role="math" localid="1663002310798" $\theta =144°$

Therefore, the angle through which loop 2 must be rotated so that the magnitude of the net field$100\text{nT}$ is $144°$.