An electron is shot into a uniform field of 6.5 G (1 G = 10⁻⁴ T) at 4.8 × 10⁶ m s⁻¹ normal to the field. With e = 1.6 × 10⁻¹⁹ C and mₑ = 9.1 × 10⁻³¹ kg, the radius of its circular orbit is about:

1. Convert the field: $B = 6.5 \text{ G} = 6.5 \times 10^{-4}$ T. 2. Radius: $r = \dfrac{mv}{qB}$. 3. Numerator: $mv = 9.1\times10^{-31} \times 4.8\times10^{6} = 4.37\times10^{-24}$. 4. Denominator: $qB = 1.6\times10^{-19} \times 6.5\times10^{-4} = 1.04\times10^{-22}$. 5. So $r = \dfrac{4.37\times10^{-24}}{1.04\times10^{-22}} \approx 4.2\times10^{-2}$ m. _Source: NCERT Class 12 Physics Ch 4 "Moving Charges and Magnetism", p.159_