Question

A long, straight wire carrying a current of 30 A is placed in an external, uniform magnetic field of 4.0 × 10⁻⁴ T parallel to the current. Find the magnitude of the resultant magnetic field at a point 2.0 cm away from the wire.

Answer 1

Given:
Uniform magnetic field, B₀ = 4.0 × 10⁻⁴ T
Magnitude of current, I = 30 A
Separation of the point from the wire, d = 0.02 m
Thus, the magnetic field due to current in the wire is given by
$B=\frac{{\mu }_{0}I}{2\mathrm{\pi }d}$

$$\begin{gathered} =\frac{2\times {10}^{-7}\times 30}{0.02} \\ =3\times {10}^{-4}\mathrm{T} \end{gathered}$$

B₀ is perpendicular to B (as shown in the figure).
∴ Resultant magnetic field
$$\begin{gathered} B_{\mathrm{net}}=\sqrt{B^2+{{\mathrm{B}}_0}^2} \\ =\sqrt{(4\times {10}^{-4}{)}^2+(3\times {10}^{-4}{)}^2} \\ =5\times {10}^{-4}\mathrm{T} \end{gathered}$$