An electron mass - 9-10-31kg- charge -1-6-10-19 C whose kinetic energy is 7-2-10-20 J is moving in a circular orbit in a magnetic field of 9- 10-5 weber-m2- The radius of the orbit is

R=√(2mK)/qB=√(2× 9×10^ 31×7.2× 10^ 20)/1.6×10^ 19× 9×10^ 5=2.5 cm ...

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Standard XIII

Physics

Circular Kinematics

An electron m...

Question

An electron (mass = $9 \times 10^{- 31} k g,$ charge $= 1.6 \times 10^{- 19}$ C ) whose kinetic energy is $7.2 \times 10^{- 20}$ J is moving in a circular orbit in a magnetic field of $9 \times 10^{- 5} w e b e r / m^{2}$ . The radius of the orbit is

1.25 cm

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2.5 cm

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12.5 cm

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25.0 cm

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Solution

The correct option is B 2.5 cm
$r = \frac{\sqrt{2 m K}}{q B} = \frac{\sqrt{2 \times 9 \times 10^{- 31} \times 7.2 \times 10^{- 20}}}{1.6 \times 10^{- 19} \times 9 \times 10^{- 5}} = 2.5 c m$

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An electron (mass = 9×10−31kg, charge =1.6×10−19 C ) whose kinetic energy is 7.2×10−20 J is moving in a circular orbit in a magnetic field of 9×10−5 weber/m2. The radius of the orbit is

The correct option is B 2.5 cmr=√2mKqB=√2×9×10−31×7.2×10−201.6×10−19×9×10−5=2.5 cm

An electron (mass = $9 \times 10^{- 31} k g,$ charge $= 1.6 \times 10^{- 19}$ C ) whose kinetic energy is $7.2 \times 10^{- 20}$ J is moving in a circular orbit in a magnetic field of $9 \times 10^{- 5} w e b e r / m^{2}$ . The radius of the orbit is

The correct option is B 2.5 cm
$r = \frac{\sqrt{2 m K}}{q B} = \frac{\sqrt{2 \times 9 \times 10^{- 31} \times 7.2 \times 10^{- 20}}}{1.6 \times 10^{- 19} \times 9 \times 10^{- 5}} = 2.5 c m$