Question

A charged particle projected in a magnetic field $B=(3\hat{i}+4\hat{j})\times {10}^{-2}T$ and the acceleration of the particle is found to be $a=(x\hat{i}+2\hat{j})m{s}^{-2}$. Then find the value of $x$.

Answer 1

Step 1: Given data

Magnetic field produced by the charged particle, $\overrightarrow{B}=(3\hat{i}+4\hat{j})\times {10}^{-2}T$

Acceleration of the particle, $\overrightarrow{a}=(x\hat{i}+2\hat{j})m{s}^{-2}$

Step 2: To find

The value of $x$.

Step 3: Formula used and calculations

We know, force on a charged particle moving perpendicular to the magnetic field is given by,

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

where $q$ is charge, $\overrightarrow{v}$ is velocity and $\overrightarrow{B}$ is magnetic field.

Since dot product of 2 perpendicular vectors is zero,

$$\begin{gathered} \overrightarrow{F}·\overrightarrow{B}=0 \\ \Rightarrow m\left(\overrightarrow{a}.\overrightarrow{B}\right)=0(\because \overrightarrow{F}=m\overrightarrow{a}) \\ \Rightarrow \overrightarrow{a}.\overrightarrow{B}=0 \end{gathered}$$

Now, on substituting the given values,

$$\begin{gathered} \overrightarrow{a}.\overrightarrow{B}=0 \\ \left(x\hat{i}+2\hat{j}\right)·\left(3\hat{i}+4\hat{j}\right)=0 \\ 3x+8=0 \\ x=-\frac{8}{3} \end{gathered}$$

Therefore, the value of $x$ is $-\frac{8}{3}$