Question

A charged particle of specific charge 0.2 C/kg has velocity of (2i + 3j) at some instant in uniform magnetic field (5i + 2j). What is the acceleration of the particle at this instant?

Answer 1

Step 1: Given data

Acceleration: In simple words, the rate of change of velocity of a thing or object with respect to time. It is a vector quantity.

A charged particle of specific charge $0.2\raisebox{1ex}{$\mathrm{C}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{kg}$}\right.$

velocity of (2i + 3j) and uniform magnetic field (5i + 2j)

Step 2: Calculations

So, now we have

$\overrightarrow{F_m}=q(\overrightarrow{V}\times \overrightarrow{B})$

And, $\overrightarrow{a}=\frac{\overrightarrow{F_m}}{m}=\frac{q}{m}(\overrightarrow{V}\times \overrightarrow{B})$

Now we need to find the acceleration.

So,

$$\begin{gathered} \overrightarrow{a}=0.2\left\{\right(2i+3j)\times (5i+2j\left)\right\} \\ =0.2\left(10i\times i+4i\times j+15j\times i+6j\times j\right) \\ =0.2\left\{4k+15\left(-k\right)\right\}\because \left(i\times i=j\times j=0\right) \end{gathered}$$

$$\begin{gathered} \overrightarrow{a}=0.2(-11k) \\ \overrightarrow{a}=-2.2m/s \end{gathered}$$

Hence, The acceleration of the particle at that instant is $-2.2m/s$.