Q183 / 200 · TS EAMCET 2002 · Physics
A proton of velocity a $(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}) \mathrm{ms}^{-1}$ enters field of magnetic induction $(2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})$ tesla, the acceleration produced in the proton in $\mathrm{ms}^{-2}$ is $\left(\right.$ Specific charge of proton $=0.96 \times 10^8 \mathrm{Ckg}^{-1}$ )
- A$28 \times 10^8(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}})$
- Correct answerB$288 \times 10^8(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})$
- C$28 \times 10^8(2 \hat{\mathbf{i}}+3 \hat{\mathbf{k}})$
- D$288 \times 10^8(\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})$
Solution
$\mathbf{V}=(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}, \mathbf{B}=(2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \mathrm{T}$
Specific charge $\frac{e}{m}=0.96 \times 10^8 \mathrm{Ckg}^{-1}$
Force acting on the proton $\mathbf{F}=q(\mathbf{V} \times \mathbf{B})$
$$
\begin{aligned}
\mathbf{V} \times \mathbf{B} & =\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
3 & 2 & 0 \\
0 & 2 & 3
\end{array}\right| \\
& =6 \hat{\mathbf{i}}-9 \hat{\mathbf{j}}+6 \hat{\mathbf{k}} \\
F & =q(6 \hat{\mathbf{i}}-9 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})
\end{aligned}
$$
From Newton's law $\mathbf{F}=$ ma
$$
m \mathbf{a}=q(6 \hat{\mathbf{i}}-9 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})
$$
$$
\begin{aligned}
\therefore \text { Acceleration } \mathbf{a} & =\frac{q}{m}(6 \hat{\mathbf{i}}-9 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}) \\
& =\frac{e}{m}(6 \hat{\mathbf{i}}-9 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}) \\
& =0.96 \times 10^8 \times 3(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \\
& =288 \times 10^8(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})
\end{aligned}
$$
Hence, the correct answer is option: 'B'.