Question: A wire carrying a${}^{\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{‐}\mathbf{A}}$current passes between the poles of a strong magnet that is perpendicular to its field and experiences a${}^{\mathbf{2}\mathbf{.}\mathbf{16}\mathbf{‐}\mathbf{N}}$force on the${}^{\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{cm}}$of wire in the field. What is the average field strength?

The current of the lightning bolt${}^{\mathbf{I}\mathbf{}\mathbf{=}\mathbf{}\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{A}}$

The magnetic force exerted is${}^{\mathbf{F}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{16}\mathbf{}\mathbf{N}}$

The length of the wire is${}^{\mathbf{L}\mathbf{}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{cm}\mathbf{}\left(\frac{1m}{100\mathrm{cm}}\right)\mathbf{}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{2}}\mathbf{}\mathbf{m}}$

The magnetic force on a current-carrying conductor can be determined using the right-hand rule-1 such that, if fingers point towards the magnetic field and the thumb is in the direction of current then the palm will not be in the direction of magnetic force.

Quantitatively magnetic force can be estimated using the equation,

$\mathbf{F}\mathbf{=}\mathbf{ILBsin}\left(\theta \right)$.................................( 1 )

The direction of the current and the magnetic field is said to be perpendicular, therefore

${}^{\mathbf{sin}\mathbf{\left(}{\mathbf{90}}^{\mathbf{°}}\mathbf{\right)}\mathbf{=}\mathbf{1}}$

Substituing the given data in the equation (1) will result in,

$$\begin{gathered} \mathbf{B}\mathbf{=}\frac{\mathbf{F}}{\mathbf{IL}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{2}\mathbf{.}\mathbf{16}}{\left(4.00\times {10}^{-2}m\right)\mathbf{\times }\left(30.0a\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{80}\mathbf{}\mathbf{T} \end{gathered}$$

Therefore, the average field strength is${}^{\mathbf{1}\mathbf{.}\mathbf{80}\mathbf{}\mathbf{T}}$.