Medium MCQ +4 / -1 PYQ · JEE Mains 2020

A particle of charge q and mass m is moving with a velocity $- v\widehat i$ (v $\ne$ 0) towards a large screen placed in the Y - Z plane at a distance d. If there is a magnetic field $\overrightarrow B = {B_0}\widehat k$ , the maximum value of v for which the particle will not hit the screen is :

  1. A ${{2qd{B_0}} \over m}$
  2. B ${{qd{B_0}} \over {3m}}$
  3. C ${{qd{B_0}} \over {2m}}$
  4. D ${{qd{B_0}} \over {m}}$ Correct answer

Solution

In uniform magnetic field particle moves in a circular path, if the radius of the circular path is 'd', particle will not hit the screen. <br><br>r = ${{mv} \over {q{B_0}}}$ <br><br>To not collide, r &lt; d <br><br>$\Rightarrow$ ${{mv} \over {q{B_0}}}$ &lt; d <br><br>$\Rightarrow$ v &lt; ${{q{B_0}d} \over m}$ <br><br>$\therefore$ v<sub>max</sub> = ${{q{B_0}d} \over m}$

About this question

Subject: Physics · Chapter: Magnetic Effects of Current · Topic: Biot-Savart Law

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