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JEE PHYSICS MCQ: MAGNETIC EFFECT OF CURRENT

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Application +4 | -1
A particle with charge 7.00μC7.00 \mu \mathrm{C} is moving with velocity v⃗=−(4×103 m s−1)j^\vec{v}=-\left(4 \times 10^3 \mathrm{~m} \mathrm{~s}^{-1}\right) \hat{j}. The magnetic force on the particle is measured to be F⃗=+(8.4×10−2 N)i^−(5.60×10−2 N)k^\vec{F}=+\left(8.4 \times 10^{-2} \mathrm{~N}\right) \hat{i}-\left(5.60 \times 10^{-2} \mathrm{~N}\right) \hat{k}.
(a) Calculate all the components of the magnetic field you can from this information.
(b) Calculate the scalar product B⃗⋅F⃗\vec{B} \cdot \vec{F}. What is the angle between B⃗\vec{B} and F⃗\vec{F}?
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Correct option: A
−2 T,−3 T;0,90∘-2 \mathrm{~T}, -3 \mathrm{~T} ; 0, 90^{\circ}
Option A (−2 T,−3 T;0,90∘-2 \text{{ T}}, -3 \text{{ T}}; 0, 90^{\circ}) is correct because the magnetic field components are calculated using the Lorentz force equation, and the scalar product of B and F is zero, indicating they are perpendicular.

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