Question 268: Magnetism and Matter

Question

The charged particle moving in a uniform magnetic field of $(3 \hat{i}+2 \hat{j}) \mathrm{T}$ has an acceleration $\left(4 \hat{i}-\frac{x}{2} \hat{j}\right) \mathrm{m} / \mathrm{s}^2$. The value of $x$ is

Accepted Answer

For a charged particle moving only under a magnetic field, the magnetic force is
$$ \vec{F} = q(\vec{v} 	imes \vec{B}) $$
So the acceleration is
$$ \vec{a} = \frac{\vec{F}}{m} = \frac{q}{m}(\vec{v} 	imes \vec{B}) $$
This means $\vec{a}$ is always perpendicular to $\vec{B}$.
So, we use the condition
$$ \vec{a} \cdot \vec{B} = 0 $$
Given,
$$ \vec{B} = (3\hat{i} + 2\hat{j}) \, 	ext{T} $$
and
$$ \vec{a} = \left(4\hat{i} - \frac{x}{2}\hat{j}ight)\, 	ext{m/s}^2 $$
Now take dot product:
$$ \vec{a} \cdot \vec{B} = \left(4\hat{i} - \frac{x}{2}\hat{j}ight)\cdot(3\hat{i} + 2\hat{j}) = 0 $$
$$ 4 \cdot 3 + \left(-\frac{x}{2}ight)\cdot 2 = 0 $$
$$ 12 - x = 0 $$
$$ x = 12 $$
Hence, the value of $x$ is
$$ \boxed{12} $$

Magnetism and Matter

The charged particle moving in a uniform magnetic field of (3 \hat{i}+2 \hat{j}) \mathrm{T} has an acceleration \left(4 \hat{i}-\frac{x}{2} \hat{j}\right) \mathrm{m} / \mathrm{s}^2. The value of x is

Integer / Numeric Answer Type