(a) A 0.750-m-long section of cable carrying current to a car starter motor makes an angle of 60º with the Earth’s ${}^{\mathbf{5}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{5}}}$field. What is the current when the wire experiences a force of ${}^{\mathbf{7}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{3}}}$? (b) If you run the wire between the poles of a strong horseshoe magnet, subjecting 5.00 cm of it to a 1.75-T field, what force is exerted on this segment of wire?

(a)

The length of the cable is${}^{{\mathbf{L}}_{\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{750}\mathbf{}\mathbf{m}}$

The angle formed by the motor is${}^{\mathbf{\theta }\mathbf{=}{\mathbf{60}}^{\mathbf{°}}}$

The earth’s magnetic field is${}^{{\mathbf{B}}_{\mathbf{1}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{}\mathbf{T}}$

The force exerted on the cable is${}^{{\mathbf{F}}_{\mathbf{1}}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\mathbf{N}}$

(b)

The length of the cable is ${}^{{\mathbf{L}}_{\mathbf{2}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{cm}\left(\frac{1m}{100\mathrm{cm}}\right)\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{}\mathbf{m}}$

The earth’s magnetic field is ${}^{{\mathbf{B}}_{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{75}\mathbf{}\mathbf{T}}$

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 )

To calculate the current flowing through the wire, we'll substitute the data in equation (1), which gives,

$$\begin{gathered} \mathbf{I}\mathbf{=}\frac{{\mathbf{F}}_{\mathbf{1}}}{{\mathbf{L}}_{\mathbf{1}}{\mathbf{B}}_{\mathbf{1}}\mathbf{sin}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}} \\ \mathbf{}\mathbf{=}\frac{\mathbf{7}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\mathbf{N}}{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{750}\mathbf{}\mathbf{m}\mathbf{)}\mathbf{\times }\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{}\mathbf{T}\mathbf{)}\mathbf{\times }\mathbf{sin}\mathbf{\left(}{\mathbf{60}}^{\mathbf{°}}\mathbf{\right)}} \\ \mathbf{}\mathbf{}\mathbf{=}\mathbf{196}\mathbf{}\mathbf{A} \end{gathered}$$

Therefore, the current flowing through the wire is${}^{\mathbf{196}\mathbf{}\mathbf{A}}$.

Further, using the equation (1) to calculate the exerted force perpendicular to the wire ($\theta ={60}^{°}$), will be such that

$$\begin{gathered} {\mathbf{F}}_{\mathbf{2}}\mathbf{=}{\mathbf{IL}}_{\mathbf{2}}{\mathbf{B}}_{\mathbf{2}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\left(5.00\times {10}^{-2}m\right)\mathbf{\times }\left(196A\right)\mathbf{\times }\left(1.75T\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{17}\mathbf{.}\mathbf{2}\mathbf{}\mathbf{N} \end{gathered}$$

Hence, the wire is subjected to a force of${}^{17.2\mathrm{N}}$.