Question

An electron (mass \[9 \times 10^{-31}\] kg, charge \[1.6 \times 10^{-19}\] C) moves at \[3 \times 10^7\] m/s in a magnetic field of \[6 \times 10^{-4}\] T perpendicular to its velocity. The radius of its circular path is:

Options

• 2.8 m = 280 cm

• 0.14 m = 14 cm

• 0.56 m = 56 cm

• 0.28 m = 28 cm

Answer 1

Using \[r = \frac{mv}{qB} = \frac{9 \times 10^{-31} \times 3 \times 10^{7}}{1.6 \times 10^{-19} \times 6 \times 10^{-4}}\]. Numerator: \[2.7 \times 10^{-23}\]; denominator: \[9.6 \times 10^{-23}\]. This gives r = 0.28 m = 28 cm.