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Physics Question on Magnetic Field

An electron having charge 1.6×10−19C1.6 \times 10^{-19} C and mass 9×10−31kg9 \times 10^{-31} kg is moving with 4×106m/s4 \times 10^{6} m / s speed in a magnetic field of 2×10−1T2 \times 10^{-1} T in a circular orbit. The force acting on an electron and the radius of circular orbit will be

A

1.28×10−14N,1.1×10−3m1.28 \times 10^{-14} N , 1.1 \times 10^{-3} m

B

1.28×1015N,1.2×10−12m1.28 \times 10^{15} N , 1.2 \times 10^{-12} m

C

1.28×10−13N,1.1×10−4m1.28 \times 10^{-13} N , 1.1 \times 10^{-4} m

D

None of the above

Answer

1.28×10−13N,1.1×10−4m1.28 \times 10^{-13} N , 1.1 \times 10^{-4} m

Explanation

Solution

Electron moves in a magnetic field (B⃗)(\vec{ B }) in a circular orbit of radius (r)(r), hence Centripetal force
== force due to magnetic field (B)(B)
mv2r=evB\frac{m v^{2}}{r}=e v B
⇒r=mveB\Rightarrow r=\frac{m v}{e B}
Given, m=9×10−31kg,m=9 \times 10^{-31} kg ,
e=1.6×10−19C,e=1.6 \times 10^{-19} C ,
v=4×106m/s,v =4 \times 10^{6} m / s ,
B=2×10−1TB=2 \times 10^{-1} T
∴r=9×10−31×4×1061.6×10−19×2×10−1\therefore r =\frac{9 \times 10^{-31} \times 4 \times 10^{6}}{1.6 \times 10^{-19} \times 2 \times 10^{-1}}
=1.1×10−4m=1.1 \times 10^{-4} m