In a circuit with one battery, connecting wires, and a $12cm$length of Nichrome wire, a compass deflection of $6°$is observed. What compass deflection would you expect in a circuit containing two batteries in a series, connecting wires and a$36cm$ length of thicker Nichrome wire (double the cross-sectional area of the thin piece)? Explain.

The length of the Nichrome wire in the first circuit is, $r_1=12\text{cm}$.

The cross-sectional area of the Nichrome wire in the first circuit is,$A_1$.

The compass deflection inthe first circuit is,${\theta }_1=6°$.

The length of the Nichrome wire in the second circuit is,$r_2=36\text{cm}$.

The cross-sectional area of the Nichrome wire in the second circuit is, $A_2=2A_1$.

When a current is supplied to an electrical circuit then the magnetic field is produced in the circuit. The magnetic field produced in a circuit relies upon the distance of the electric field from the circuit, and the value of the current supplied.

If the current supplied to conducting material is increased then the strength of the magnetic field produced due to current also increases.

The current flowing through the circuit is calculated using formula,

$$\begin{gathered} I=\left|q\right|nAuE \\ I\propto A \end{gathered}$$

So, the current in the second circuit will be,

$I_2=2I_1$

The formula for the magnetic field made by the current in the first circuit is given by,

$B_1=\frac{{\mu }_0}{4\pi }\frac{2I_1}{r_1}$

Then, the deflection angle in the first circuit is given by,

$$\begin{gathered} \frac{B_1}{B_{\text{Earth}}}=\tan {\theta }_1 \\ \frac{B_1}{\tan {\theta }_1}=B_{\text{Earth}}...\left(1\right) \end{gathered}$$

Similarly, the magnetic field made by the current in the second circuit is given by,

$B_2=\frac{{\mu }_0}{4\pi }\frac{2I_2}{r_2}$

Then, the deflection angle in the second circuit is given by,

$$\begin{gathered} \frac{B_2}{B_{\text{Earth}}}=\tan {\theta }_2 \\ \frac{B_2}{\tan {\theta }_2}=B_{\text{Earth}} \end{gathered}$$

From equation (1),

$$\begin{gathered} \frac{B_2}{\tan {\theta }_2}=\frac{B_1}{\tan {\theta }_1} \\ \tan {\theta }_2=\frac{B_2}{B_1}\tan {\theta }_1 \\ \tan {\theta }_2=\frac{\frac{{\mu }_0}{4\pi }\frac{2I_2}{r_2}}{\frac{{\mu }_0}{4\pi }\frac{2I_1}{r_1}}\tan {\theta }_1 \\ \tan {\theta }_2=\frac{I_2{r}_1}{I_1{r}_2}\tan {\theta }_1 \end{gathered}$$

Putting $I_2=2I_1$in expression,

$$\begin{gathered} \tan {\theta }_2=\frac{2I_1{r}_1}{I_1{r}_2}\tan {\theta }_1 \\ \tan {\theta }_2=\frac{r_1}{r_2}\tan {\theta }_1 \end{gathered}$$

Putting all the values,

$$\begin{gathered} \tan {\theta }_2=\frac{12\text{cm}}{36\text{cm}}\tan \left(6°\right) \\ \tan {\theta }_2=0.0350 \\ {\theta }_2=2° \end{gathered}$$

Hence, the compass deflection in a circuit containing two batteries in a series is $2°$.