Question

Electrons move at right angle to a magnetic field of $1.5\times {10}^{-2}$ Tesla with a speed of $6\times {10}^7$ m/s. If the specific charge of the electron is $1.7\times {10}^{11}$ C/kg, then the radius of the circular path will be?

• A. 2.9 cm
• B. 3.9 cm
• C. 2.35 cm
• D. 3 cm

Answer 1

The correct option is C 2.35 cm

$r=\frac{mv}{qB}\Rightarrow \frac{v}{\left(q/m\right).B}=\frac{6\times {10}^7}{1.7\times {10}^{11}\times 1.5\times {10}^{-2}}=2.35\times {10}^{-2}m=2.35cm$