Question

An electron moves through a uniform magnetic field given by$\overrightarrow{\mathbf{B}}$=B_{xlocalid="1663949077851" $\hat{\mathbf{i}}\mathbf{+}\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{0}$}B_{xlocalid="1663949086294" $\mathbf{)}\hat{\mathbf{j}}$}. At a particular instant, the electron has velocity$\overrightarrow{\mathbf{v}}$= (localid="1663949095061" $\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}$) and the magnetic force acting on it islocalid="1663949102219" $\left(6.4\times {10}^{-19}N\right)\hat{\mathbf{k}}\mathbf{.}$Find B_x.

Short answer

The value of$B_x\text{is - 2.0T.}$

Step-by-step solution

Step 1: Given

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

The force on the electron due to magnetic field,$\stackrel{\rightharpoonup }{F_B}=(6.4\times {10}^{-19}N)\stackrel{\rightharpoonup }{k}$

The magnetic field,$\stackrel{\rightharpoonup }{B}=(B_x)\hat{i}+(3.0B_x)\hat{j}$

Determining the concept

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.

$F_B=qv\times B$

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

Determining the value of 

The magnetic force experienced bytheelectron is,

$F_B=e\left[\overrightarrow{v}\times \overrightarrow{B}\right]$

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

localid="1662384499354" $\overrightarrow{v}\times \overrightarrow{B}$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|$

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

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

Hence,

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

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

$B_x\text{= - 2.0T}$

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