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Question: An electron beam passes through a magnetic field \(2 \times 10^{- 3}weber/m^{2}\)and an electric fie...

An electron beam passes through a magnetic field 2×10−3weber/m22 \times 10^{- 3}weber/m^{2}and an electric field 3.4×10−43.4 \times 10^{- 4}volts/m both acting simultaneously. If the path of electrons remains undeviated, calculate the speed of the electrons. If the electric field is removed, what will be radius of the electron path?

A

2.4×10−2m2.4 \times 10^{- 2}m

B

3.86×10−4m3.86 \times 10^{- 4}m

C

4.78×10−2m4.78 \times 10^{- 2}m

D

5×10−2m5 \times 10^{- 2}m

Answer

4.78×10−2m4.78 \times 10^{- 2}m

Explanation

Solution

For the undeviated path qE=qvBqE = qvB

or v=EB=3.4×1042×10−3=1.7×107m/sv = \frac{E}{B} = \frac{3.4 \times 10^{4}}{2 \times 10^{- 3}} = 1.7 \times 10^{7}m/s

When the electric field is removed, then Bev=mv2rBev = \frac{mv^{2}}{r} or r=mvBer = \frac{mv}{Be}

= 9×10−31×(1.7×107)(2×10−3)×(1.6×10−19)\frac{9 \times 10^{- 31} \times \left( 1.7 \times 10^{7} \right)}{\left( 2 \times 10^{- 3} \right) \times \left( 1.6 \times 10^{- 19} \right)}

(∵m=9×10−31kg)\left( \because m = 9 \times 10^{- 31}kg \right)

= 4.78×10−2m4.78 \times 10^{- 2}m