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)

6

Answer 2

The correct option is (B) : 6
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 ⊥ to each other
Hence , $\vec{a}.\vec{B}=0$
$(\alpha\hat{i}-4\hat{j}).(2\hat{i}+3\hat{j})=0$
$\alpha(2)+(-4)(3)=0$
$\alpha=\frac{12}{2}$
$⇒ \alpha =6$