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  • I have a doubt, kindly clarify. A particle having the same charge as of electron moves in a circular path of radius 0.5cm under the influence of a magnetic field of 0.5T. If an electric field of 100V/m makes it to move in a straight path, then the mass of

A particle having the same charge as of electron moves in a circular path of radius 0.5cm under the influence of a magnetic field of 0.5T. If an electric field of 100V/m makes it to move in a straight path, then the mass of the particle is (Given charge of electron = 1.6 X 10-19C)

 

  • Option 1)

    9.1 \times 10^{-31}kg

  • Option 2)

    1.6 \times 10^{-19}kg

  • Option 3)

    1.6 \times 10^{-27}kg

  • Option 4)

    2.0 \times 10^{-24}kg

Answers (1)

best_answer

 

Radius of charged particle -

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

-

\frac{mv^{2}}{R}=qvB

mv=qBR

Path is straight

so,

qE=qvB

E=vB

\therefore  m\frac{E}{B}=qBR

m=\frac{qB^{2}R}{E}

      =\frac{1.6\times 10^{-19}\times (0.5)^{2}\times 0.5\times 10^{-2}}{100}

     =2\times 10^{-24}\: kg


Option 1)

9.1 \times 10^{-31}kg

Option 2)

1.6 \times 10^{-19}kg

Option 3)

1.6 \times 10^{-27}kg

Option 4)

2.0 \times 10^{-24}kg
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