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Physics Question on Magnetic Field

An electron having charge 1.6×10−191.6\times {{10}^{-19}} and mass 9×10−31kg9\times {{10}^{-31}}kg is moving with 4×106ms−14\times {{10}^{6}}m{{s}^{-1}} speed in a magnetic field 2×10−12\times {{10}^{-1}} tesla in a circular orbit. The force acting on electron and the radius of the circular orbit will be:

A

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

B

1.28×10−13  N, 1.1 10−3m1.28\times {{10}^{-13~~}}N,\text{ }1.1\text{ }{{10}^{-3}}m

C

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

D

12.8×10−13 N, 1.1×10−4m12.8\times {{10}^{-13~}}N,\text{ }1.1\times {{10}^{-4}}m

Answer

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

Explanation

Solution

From the relation F=qvB=1.6×10−19×4×106×2×10−1F=qvB=1.6\times {{10}^{-19}}\times 4\times {{10}^{6}}\times 2\times {{10}^{-1}} =1.28×10−13N=1.28\times {{10}^{-13}}N Radius of circular orbit is given by r=mυqB=9×10−31×4×1061.6×10−19×2×10−1r=\frac{m\upsilon }{qB}=\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