91Ó°ÊÓ

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

$\mathbf{B}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}\mathbf{l}}{\mathbf{2}\mathbf{Ï„Ï„°ù}}$

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.

$\mathbf{B}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}\mathbf{l}}{\mathbf{2}\mathbf{Ï„Ï„°ù}}\hat{\mathbf{k}}$ …(¾±)

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

$\begin{array}{r}\mathrm{F}=\mathrm{q}(\stackrel{→}{\mathrm{V}}×\stackrel{→}{\mathrm{B}})\\ =−\mathrm{e}(\stackrel{→}{\mathrm{V}}×\stackrel{→}{\mathrm{B}})\end{array}$ …(¾±¾±)

The magnitude of the magnetic field is,

$\begin{array}{r}\mathrm{B}=\frac{{\mathrm{μ}}_0\left(9.00\mathrm{A}\right)}{2\mathrm{π}\left(0.200\mathrm{m}\right)}\\ =9.00×{10}^{−6}\mathrm{T}\end{array}$

So, by equation (i),

$\stackrel{→}{\mathrm{B}}=−9.00×{10}^{−6}\mathrm{T}\widehat{\mathrm{k}}$

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

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

$\stackrel{→}{\mathrm{F}}=−4.32×{10}^{−20}\mathrm{N}\widehat{\mathrm{i}}−7.20×{10}^{−20}\mathrm{N}\widehat{\mathrm{j}}$

Now, calculate the magnitude,

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

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