Question

(a) Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a $\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{‐}\mathbf{\mu C}$ charge and flies due west at a speed of 660 m/s over the Earth’s south magnetic pole, where the $\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{5}}\mathbf{‐}\mathbf{T}$magnetic field points straight up. What are the direction and the magnitude of the magnetic force on the plane? (b) Discuss whether the value obtained in part (a) implies this is a significant or negligible effect.

Short answer

(a) The magnitude of the force is $2.64\times {10}^{‐8}\mathrm{N}$and it points towards the earth.

(b) The effect of this force is insignificant.

Step-by-step solution

Given information

The charge on the supersonic jet is \({\bf{q = 0}}.5{\bf{00}}\;{\bf{\mu C}}\left( {\frac{{{\bf{1}}{{\bf{0}}^{{\bf{ - 6}}}}\;{\bf{C}}}}{{{\bf{1}}\;{\bf{\mu C}}}}} \right) = 5.00 \times {\bf{1}}{{\bf{0}}^{{\bf{ - }}7}}\;{\bf{C}}\)

The magnetic field strength of the earth’s south pole is$\mathrm{B}=8.00\times {10}^{‐5}\mathrm{T}$

The speed of the supersonic jet is$\mathrm{v}=660\mathrm{m}/\mathrm{s}$

Step 2:Definition of magnetic force

The magnetic force is a consequence of the electromagnetic force, one of the four fundamental forces of nature, and is caused by the motion of charges. Two objects containing a charge with the same direction of motion have a magnetic attraction force between them. Similarly, objects with charges moving in opposite directions have a repulsive force between them.

Step 3:Determining the magnetic force exerted on the plane

(a)

To determine the magnetic force exerted on the plane, we will use the equation,

$\mathrm{F}=\mathrm{qvBsin}\left(\mathrm{\theta }\right)$…………………………(1)

where $\theta$is the angle between the direction in which the plane moves, to the west, and the direction of the earth's magnetic field, to the north. This angle will be .

The direction of the force will be determined by the right-hand rule. By pointing the thumb to the left, the direction of the velocity, and pointing the fingers upwards, the direction of the earth's magnetic field, the force will be towards the earth.

Substituting the given values in the above expression will give,

$$\begin{gathered} \mathbf{F}\mathbf{=}\mathbf{(}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{7}}\mathbf{}\mathbf{C}\mathbf{)}\mathbf{\times }\mathbf{(}\mathbf{660}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{)}\mathbf{\times }\mathbf{(}\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{5}}\mathbf{}\mathbf{T}\mathbf{)}\mathbf{\times }\mathbf{sin}\left({90}^{\circ }\right) \\ \mathbf{}\mathbf{}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{64}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{8}}\mathbf{}\mathbf{N} \end{gathered}$$

Therefore, the magnitude of the force is $\mathbf{2}\mathbf{.}\mathbf{64}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{8}}\mathbf{}\mathbf{N}$and it points toward the earth.

(b)

Considering the mass of the plane, we can surely say that the effect of this force is insignificant.