An electron (mass 9 × 10⁻³¹ kg, charge 1.6 × 10⁻¹⁹ C) moves at 3 × 10⁷ m s⁻¹ perpendicular to a field of 6 × 10⁻⁴ T. The radius of its circular path is approximately:

1. Radius of circular motion: $r = \dfrac{mv}{qB}$. 2. Numerator: $mv = 9\times10^{-31} \times 3\times10^{7} = 2.7\times10^{-23}$. 3. Denominator: $qB = 1.6\times10^{-19} \times 6\times10^{-4} = 9.6\times10^{-23}$. 4. So $r = \dfrac{2.7\times10^{-23}}{9.6\times10^{-23}} = 0.28$ m. 5. That is $r = 28$ cm. _Source: NCERT Class 12 Physics Ch 4 "Moving Charges and Magnetism", p.139_