First slide

Force on a charged particle moving in a magnetic field

Question

An electron $(mass = 9 .1 \times 10^{- 31} kg, charge = 1 .6 \times 10^{- 19} C)$ experiences no deflection if subjected to an electric field of $3 .2 \times 10^{5} V / m$ and a magnetic fields of $0.2 μ T$ Both the fields are normal to the path of electron and to each other. If the electric field is removed, then the electron will revolve in an orbit of radius

Moderate

A

45 m
B

4.5 m
C

0.45 m
D

0.045 m

Solution

For no deflection in mutually perpendicular electric an magnetic field, $v = \frac{E}{B} = \frac{3 .2 \times 10^{3}}{0.2} = 1 .6 \times 10^{4} m / s .$
If electric field is removed, then due to only magnetic field, radius of the path described by electron is
$r = \frac{mv}{qB} = \frac{9 .1 \times 10^{- 31} \times 1 .6 \times 10^{4}}{1 .6 \times 10^{- 19} \times 0.2 \times 10^{- 6}} = 0 .45 m$

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