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Physics Question on Moving charges and magnetism

A particle having the same charge as of electron moves in a circular path of radius 0.5 cm0.5\, cm under the influence of a magnetic field of 0.5 T0.5\, T. If an electric field of 100 V/m100\, V/m makes it to move in a straight path, then the mass of the particle is (Given charge of electron = 1.6×10−19C)1.6 \times 10^{-19} C)

A

2.0×10−24  kg2.0 \times 10^{-24} \; kg

B

1.6×10−19  kg1.6 \times 10^{-19} \; kg

C

1.6×10−27  kg1.6 \times 10^{-27} \; kg

D

9.1×10−31  kg9.1 \times 10^{-31} \; kg

Answer

2.0×10−24  kg2.0 \times 10^{-24} \; kg

Explanation

Solution

mv2R=qvB\frac{mv^{2}}{R} = qvB
mv=qBRmv =qBR ....(i)
Path is straight line
it qE=qvBqE = qvB
E=vBE = vB ....(ii)
From equation (i) & (ii)
m=qB2REm = \frac{qB^{2}R}{E}
m=2.0×10−24kgm = 2.0 \times10^{-24} kg