Question

An electron is moving in the vicinity of a long, straight wire that lies along the x-axis. The wire has a constant current of 9.00 A in the -x-direction. At an instant when the electron is at point (0, 0.200 m, 0) and the electron’s velocity is $\overrightarrow{\mathbf{v}}\mathbf{=}\left(5.00\times {10}^{4}m/s\right)\widehat{\mathbf{i}}\mathbf{-}\left(3.00\times {10}^{4}m/s\right)\widehat{\mathbf{j}}$, what is the force that the wire exerts on the electron? Express the force in terms of unit vectors, and calculate its magnitude.

Short answer

The magnitude of force the wire exerts on the electron is$8.40\times {10}^{-20}\mathrm{N}$.

Step-by-step solution

(a) Concept of Magnetic field due to a current carrying wire.

The magnitude of the magnetic field due to current carrying wire is,

$\mathbf{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{l}}{\mathbf{2}\mathbf{\tau \tau r}}$

Here, the current is running in –x direction, the electron is travelling in the y-direction then the magnetic field, from the right hand rule, is directed along the –z direction.

$\mathbf{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{l}}{\mathbf{2}\mathbf{\tau \tau r}}\hat{\mathbf{k}}$ …(i)

(b) Determination of the magnitude of force the wire exerts on the electron.

Lorentz force on a moving charge under the influence of magnetic field is,

$\begin{array}{r}\mathrm{F}=\mathrm{q}(\overrightarrow{\mathrm{V}}\times \overrightarrow{\mathrm{B}})\\ =-\mathrm{e}(\overrightarrow{\mathrm{V}}\times \overrightarrow{\mathrm{B}})\end{array}$ …(ii)

The magnitude of the magnetic field is,

$\begin{array}{r}\mathrm{B}=\frac{{\mu }_0\left(9.00\mathrm{A}\right)}{2\pi \left(0.200\mathrm{m}\right)}\\ =9.00\times {10}^{-6}\mathrm{T}\end{array}$

So, by equation (i),

$\overrightarrow{\mathrm{B}}=-9.00\times {10}^{-6}\mathrm{T}\widehat{\mathrm{k}}$

Calculate the force from the equation (ii),localid="1668339345352" $\begin{array}{r}\overrightarrow{\mathrm{F}}=-\mathrm{e}\left(5.00\times {10}^4\mathrm{m}/\mathrm{s}\widehat{\mathrm{i}}-3.00\times {10}^4\mathrm{m}/\mathrm{s}\widehat{\mathrm{j}}\right)\times \left(-9.00\times {10}^{-6}\mathrm{T}\widehat{\mathrm{k}}\right)\\ =\left(9\times {10}^{-2}\mathrm{e}T.\mathrm{m}/\mathrm{s}\right)(5\widehat{i}-3\widehat{j})\times \widehat{k}\end{array}$

Use, the vector identities to find the final vector of the force experienced.

$\overrightarrow{\mathrm{F}}=-4.32\times {10}^{-20}\mathrm{N}\widehat{\mathrm{i}}-7.20\times {10}^{-20}\mathrm{N}\widehat{\mathrm{j}}$

Now, calculate the magnitude,

$\begin{array}{r}\mathrm{F}=\sqrt{{\mathrm{F}}_{\mathrm{x}}^2+{\mathrm{F}}_{\mathrm{y}}^2+{\mathrm{F}}_{\mathrm{z}}^2}\\ =\sqrt{{\left(-4.32\times {10}^{-20}\mathrm{N}\right)}^2+{\left(-7.20\times {10}^{-20}\mathrm{N}\right)}^2}\\ =8.40\times {10}^{-20}\mathrm{N}\end{array}$

Thus, the magnitude of the force is $8.40\times {10}^{-20}\mathrm{N}$.