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JEE PHYSICS MCQ: MAGNETIC EFFECT OF CURRENT

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An observer's diary reads as follows; "a charged particle is projected in a magnetic field of and particle's acceleration is found to be (xi^+7.0j^)×10−6 m/(x \hat{i}+7.0 \hat{j}) \times 10^{-6} \mathrm{~m} / s2\mathbf{s}^2. Find the value of x\mathbf{x}.
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Correct option: A
x=3.0x=3.0
Option A (x=3.0x=3.0) is correct because the force due to the magnetic field is perpendicular to the magnetic field itself, allowing us to find x.

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