A proton moves with velocity v = 3 × 10⁶ i m/s in a magnetic field B = 0.5 k T — JEE Physics
A proton moves with velocity $\vec{v} = 3 \times 10^6\ \hat{i}$ m/s in a magnetic field $\vec{B} = 0.5\ \hat{k}$ T. Find the magnetic force on it. ($e = 1.6 \times 10^{-19}$ C)
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 30d ago
▲ 30
$$\vec{F} = q(\vec{v} \times \vec{B}) = 1.6\times10^{-19}(3\times10^6 \hat{i} \times 0.5\hat{k})$$
$$\hat{i} \times \hat{k} = -\hat{j}$$
$$\vec{F} = 1.6\times10^{-19} \times 1.5\times10^6 (-\hat{j}) = -2.4\times10^{-13}\ \hat{j} N$$
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Discussion (3)
DM
Underrated solution. The way you set it up makes it almost obvious.
MT
Does this approach generalise to the JEE Advanced version of this question?
VA
Nicely solved. For the exam, write the final boxed answer clearly to avoid losing marks.