JEE PYQ: Magnetic Effects Of Current Question 20
Question 20 - 2020 (06 Sep 2020 Shift 1)
A particle of charge $q$ and mass $m$ is moving with a velocity $-v\hat{i}$ $(v \neq 0)$ towards a large screen placed in the Y-Z plane at a distance $d$. If there is a magnetic field $\vec{B} = B_0 \hat{k}$, the minimum value of $v$ for which the particle will not hit the screen is:
(1) $\frac{qdB_0}{3m}$
(2) $\frac{2qdB_0}{m}$
(3) $\frac{qdB_0}{m}$
(4) $\frac{qdB_0}{2m}$
Show Answer
Answer: (3) $\frac{qdB_0}{m}$
Solution
Radius $r = \frac{mv}{qB_0}$. For not hitting: $r \leq d$, so $v_{\min} = \frac{qdB_0}{m}$.
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