Question

Question 14 20 pt A proton moves with a velocity of v = (2 i-5j+k)m/s in a region in which the magnetic field is B = (i+2j-2 k) T. What is the the magnitude and direction of the magnetic force this particle experiences? Upload Choose a File < Previous o e

Answer 1

Sol: Here

ū = (21 – 5j + k)m/s

B = (i + 21 - 2k)T

Charge $q=1.6\times 10^{-19}C$

Hence force

$\vec{F}=q(\vec{v}\times \vec{B} )$

$\vec{F}=1.6\times 10^{-19}\times \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -5 &1 \\ 1 &2 & -2 \end{bmatrix}=1.6\times 10^{-19}\times (8\hat{i}+5\hat{j}+9\hat{k})=(12.8\times 10^{-19}\hat{i}+8\times 10^{-19}\hat{j}+14.4\times 10^{-19}\hat{k})N$

Hence Magnitude

$F=\sqrt{(12.8\times 10^{-19})^{2}+(8\times 10^{-19})^{2}+(14.4\times 10^{-19})^{2}}=20.86\times 10^{-19}N$

And the direction will be perpendicular to the plane in which vector v and vecto B lie. According to right hand thum rule.