Question

If magnetic force on a charge is given by q( v × B ) and a charged particle is projected in a magnetic field (2 i ^ +2 j ^ ​ +2 k ^ ) tesla. The acceleration of the particle at an instant is (x i ^ +2 j ^ ​ −6 k ^ )m/s 2 . Value of x is

Answer 1

4

Answer 2

The magnetic force on a charged particle is given by $\vec{F} = q(\vec{v} \times \vec{B})$. According to Newton's second law, the force is also given by $\vec{F} = m\vec{a}$. Therefore, we can write $m\vec{a} = q(\vec{v} \times \vec{B})$. This implies that the acceleration of the particle is $\vec{a} = \frac{q}{m}(\vec{v} \times \vec{B})$.

A fundamental property of the cross product is that the resulting vector ($\vec{v} \times \vec{B}$) is always perpendicular to both original vectors ($\vec{v}$ and $\vec{B}$). Since $\vec{a}$ is in the same direction as $(\vec{v} \times \vec{B})$, it means that the acceleration vector $\vec{a}$ must be perpendicular to the magnetic field vector $\vec{B}$.

For two vectors to be perpendicular, their dot product must be zero. So, $\vec{a} \cdot \vec{B} = 0$.

Given:

Magnetic field $\vec{B} = (2\hat{i} + 2\hat{j} + 2\hat{k})$ T Acceleration $\vec{a} = (x\hat{i} + 2\hat{j} - 6\hat{k})$ m/s$^2$

Now, calculate the dot product of $\vec{a}$ and $\vec{B}$:

$\vec{a} \cdot \vec{B} = (x\hat{i} + 2\hat{j} - 6\hat{k}) \cdot (2\hat{i} + 2\hat{j} + 2\hat{k})$ $\vec{a} \cdot \vec{B} = (x)(2) + (2)(2) + (-6)(2)$ $\vec{a} \cdot \vec{B} = 2x + 4 - 12$ $\vec{a} \cdot \vec{B} = 2x - 8$

Since $\vec{a} \cdot \vec{B} = 0$:

$2x - 8 = 0$ $2x = 8$ $x = 4$

The value of x is 4.