The charged particle moving in a uniform magnetic field of (3i + 2j) T has an acceleration (4i - x/2j) m/s². The value of x is Numerical answer type. Work it out, then open the answer.

Answer: 12. Magnetic force, hence acceleration, is perpendicular to B: a · B = 0. (4)(3) + (-x/2)(2) = 0 ⇒ 12 - x = 0 x = 12

The charged particle moving in a uniform magnetic field of (3i + 2j)… | JEE Main 2026 (5 Apr, Shift 1)

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Moving Charge in a Magnetic Field

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

Numerical answer type. Work it out, then open the answer.

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Answer: 12

Magnetic force, hence acceleration, is perpendicular to $\vec{B}$ : $\vec{a} \cdot \vec{B} = 0$ .
$(4)(3) + \left(-\frac{x}{2}\right)(2) = 0 \Rightarrow 12 - x = 0$
$x = 12$

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