Question

A charge particle is moving in a uniform magnetic field $(2\hat{i}+3\hat{j})T$. If it has an acceleration of $(\alpha \hat{i}-4\hat{j})m/s^2$, then the value of $\alpha$ will be

• A. 3
• B. 6
• C. 12
• D. 2

Answer 1

Answer: (B)

As $\vec{F}=q(\vec{v}\times \vec{B})$
$\vec{a}=\frac{q}{m}(\vec{v}\times \vec{B})$
So, $\vec{a}\&\vec{B}$ are $\perp$ to each other
Hence, $\vec{a}\cdot \vec{B}=0$
$(\alpha \hat{i}-4\hat{j})\cdot (2\hat{i}+3\hat{j})=0$
$\alpha (2)+(-4)(3)=0$
$\alpha =\frac{12}{2}\Rightarrow \alpha =6$