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  • Solve this problem - Magnetic Effects of Current and Magnetism - JEE Main

A particle of mass M and  charge  Q moving with velocity \vec{v} describes a circular path of radius R when subjected to a uniform transverse magnetic field of induction B . The work done by the field when the particle completes one full circle is

  • Option 1)

    \left ( \frac{Mv^{2}}{R} \right )2\pi R

  • Option 2)

    zero

  • Option 3)

    BQ\, \, 2\pi R

  • Option 4)

    BQv\, \, 2\pi R

 

Answers (1)

As we discussed in concept

Direction of Force (Right hand screw rule) -

The force  \underset{F}{\rightarrow} is always perpendicular to both \underset{V}{\rightarrow} and \underset{B}{\rightarrow} whereas \underset{V}{\rightarrow} and \underset{B}{\rightarrow} may or may not be perpendicular to each other 

- wherein

 

 

 

Force is always perpendicular to displacement hence Work done by the magnetic field = zero.


Option 1)

\left ( \frac{Mv^{2}}{R} \right )2\pi R

Incorrect option

Option 2)

zero

Correct option

Option 3)

BQ\, \, 2\pi R

Incorrect option

Option 4)

BQv\, \, 2\pi R

Incorrect option

Posted by

Sabhrant Ambastha

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