(a) What is the angle between a wire carrying an ${}^{\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{‐}\mathbf{A}}$current and the${}^{\mathbf{1}\mathbf{.}\mathbf{20}\mathbf{‐}\mathbf{T}}$field it is in if ${}^{\mathbf{50}\mathbf{.}\mathbf{0}}$cm of the wire experiences a magnetic force of ${}^{\mathbf{2}\mathbf{.}\mathbf{40}}$N? (b) What is the force on the wire if it is rotated to make an angle of 90º with the field?

The current of the lightning bolt${}^{\mathbf{I}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{A}}$

The earth’s magnetic field is${}^{\mathbf{B}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{20}\mathbf{}\mathbf{T}}$

The diameter of the tube is${}^{\mathbf{L}\mathbf{=}\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}\left(\frac{1m}{100\mathrm{cm}}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{m}}$

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 )

(a)

Rearrangement of equation (1) and substituting the given values in it will result in,

$$\begin{gathered} \mathbf{sin}\left(\theta \right)\mathbf{=}\frac{\mathbf{F}}{\mathbf{ILB}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{2}\mathbf{.}\mathbf{40}\mathbf{}\mathbf{N}}{\left(0.50m\right)\mathbf{\times }\left(8.00A\right)\mathbf{\times }\left(1.20T\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}} \end{gathered}$$

This can be simplified to give the value of angle,

$$\begin{gathered} \mathbf{\theta }\mathbf{=}\mathbf{arcsin}\left(\frac{1}{2}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}{\mathbf{30}}^{\mathbf{°}} \end{gathered}$$

Hence, the angle between the current in the wire and the magnetic field is${}^{{\mathbf{30}}^{\mathbf{°}}}$.

(b)

Now, to determine the exerted force on the wire, when the angle is${}^{{\mathbf{90}}^{\mathbf{°}}}$with the field, we use equation (1), which gives,

$$\begin{gathered} \mathbf{F}\mathbf{=}\mathbf{ILBsin}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathbf{} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\left(0.50m\right)\mathbf{\times }\left(8.00A\right)\mathbf{\times }\left(1.20T\right)\mathbf{\times }\mathbf{sin}\left({90}^{°}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{80}\mathbf{}\mathbf{N} \end{gathered}$$

Therefore, the required force is${}^{\mathbf{4}\mathbf{.}\mathbf{80}\mathbf{}\mathbf{N}}$ .