Question

$A$ proton of velocity $(3\overline{i}+2\overline{j})$ $m{s}^{-1}$ enters a field of magnetic induction $(2\overline{i}+3\overline{k})T$. The acceleration produced in the proton is (specific charge of proton $=0.96\times {10}^{8}Ck{g}^{-1};F=q(\overrightarrow{v}\times \overrightarrow{B})$)

• A. $0.96\times {10}^8(6\overrightarrow{i}+4\overrightarrow{k}+9\overrightarrow{j})$
• B. $0.96\times {10}^8(6\overrightarrow{i}-9\overrightarrow{j}-4\overrightarrow{k})$
• C. $0.96\times {10}^8(\overrightarrow{i}-\overrightarrow{j}-\overrightarrow{k})$
• D. $0.96\times {10}^8(5\overrightarrow{i}-9\overrightarrow{j}-4\overrightarrow{k})$

Answer 1

Answer: (B)

$0.96\times {10}^8(6\overrightarrow{i}-9\overrightarrow{j}-4\overrightarrow{k})$

$F=q(\overrightarrow{v}\times \overrightarrow{B})ora=\frac{q}{m}(\overrightarrow{v}\times \overrightarrow{B})=(0.96\times {10}^8)\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 2& 0\\ 2& 0& 3\end{array}\right|=(0.96\times {10}^8)(6\hat{i}-9\hat{j}-4\hat{k})$