91Ó°ÊÓ

The velocity of electron, $\stackrel{â‡\P Ä}{v}=(2.0)\hat{i}+(4.0)\hat{j}m/s$

The force on the electron due to magnetic field,$\stackrel{â‡\P Ä}{F_B}=(6.4×{10}^{-19}N)\stackrel{â‡\P Ä}{k}$

The magnetic field,$\stackrel{â‡\P Ä}{B}=(B_x)\hat{i}+(3.0B_x)\hat{j}$

Findthe value of${\mathbf{B}}_{\mathbf{x}}$on an electron using the formula for magnetic force in terms of charge, magnetic field strength, and velocity of an electron.

The force on moving charge due to magnetic field.

$F_B=qv×B$

Where, F_Bis magnetic force, v is velocity, B is magnetic field, q is charge of particle.

The magnetic force experienced bytheelectron is,

$F_B=e\left[\stackrel{→}{v}×\stackrel{→}{B}\right]$

$(6.4×{10}^{-19}N)\hat{k}=(-1.6×{10}^{-19}C)[((2.0)\hat{i}+(4.0)\hat{j})×((B_x)\hat{i}+(3.0B_x)\hat{j}\left)\right]............\left(1\right)$

localid="1662384499354" $\stackrel{→}{v}×\stackrel{→}{B}$localid="1662384502905" $=$$\left|\begin{array}{c}\hat{i}\\ \text{2.0}\\ {\text{B}}_{\text{x}}\end{array}\begin{array}{c}\hat{j}\\ \text{4.0}\\ {\text{3.0B}}_{\text{x}}\end{array}\begin{array}{c}\hat{k}\\ \text{0}\\ \text{0}\end{array}\right|$

$\stackrel{→}{v}×\stackrel{→}{B}=$localid="1662354923381" $\left[\right((2.0)(3.0B_x))-(4.0B_x\left)\right]\hat{k}$

$\stackrel{→}{v}×\stackrel{→}{B}=$localid="1662355745585" $\left(\left({\text{2.0B}}_{\text{x}}\right)T_{M/S}\right)$$\hat{k}$

Hence,

$\left(6.4×{10}^{-19}N\right)\hat{k}=\left(-1.6×{10}^{-19}C\right)\left(2.0B_X\right)\hat{k}$

$B_x=\frac{\left(6.4×{10}^{-19}\text{N}\right)}{\left(-1.6×{10}^{-19}\text{C}\right)\left(2.0\right)}$

$B_x\text{= - 2.0T}$

Hence, the value of magnetic field $\left(B_x\right)$is $\text{- 2.0T}$.