A particle of mass m and charge q moves with a constant velocity v along the positive x direction. It enters a region containing a uniform magnetic field B directed along the negative z direction, extending from x = a to x = b. The minimum value of v required so that the particle can just enter the region  x>b  is

 

  • Option 1)

    \frac{qbB}{m}

  • Option 2)

    \frac{q\left ( b-a \right )B}{m}

  • Option 3)

    \frac{qaB}{m}

  • Option 4)

    \frac{q\left ( b+a \right )B}{2m}

 

Answers (1)
V Vakul

As we learnt in 

Radius of charged particle -

r=\frac{mv}{qB}=\frac{P}{qB}=\frac{\sqrt{}2mk}{qB}=\frac{1}{B}\sqrt{}\frac{2mV}{q}

-

For particle to enter x > b

Radius \geq (b - a)

\Rightarrow \frac{mv}{qB}\geq b-a   or       v \geq \frac{qB(b-a)}{m}

v^{min}=\frac{qB(b-a)}{m}


Option 1)

\frac{qbB}{m}

incorrect

Option 2)

\frac{q\left ( b-a \right )B}{m}

correct

Option 3)

\frac{qaB}{m}

incorrect

Option 4)

\frac{q\left ( b+a \right )B}{2m}

incorrect

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