A charge particle of 2 $$\mu$$C accelerated by a potential difference of 100 V enters a region of uniform magnetic field of magnitude 4 mT at right angle to the direction of field. The charge particle completes semicircle of radius 3 cm inside magnetic field. The mass of the charge particle is _____ $$\times 10^{-18}$$ kg.

We have a charge $$q = 2 \mu C = 2 \times 10^{-6}$$ C accelerated through a potential difference $$V = 100$$ V, entering a magnetic field $$B = 4$$ mT $$= 4 \times 10^{-3}$$ T where it traces a semicircle of radius $$r = 3$$ cm $$= 0.03$$ m.

The kinetic energy gained by the charge after acceleration is

$$KE = qV = \frac{1}{2}mv^2$$

Now, the radius of circular motion in a magnetic field gives us

$$r = \frac{mv}{qB}$$

So $$mv = qBr$$. Substituting into the kinetic energy equation,

$$qV = \frac{(mv)^2}{2m} = \frac{(qBr)^2}{2m}$$

Solving for mass,

$$m = \frac{q^2B^2r^2}{2qV} = \frac{qB^2r^2}{2V}$$ $$m = \frac{2 \times 10^{-6} \times (4 \times 10^{-3})^2 \times (0.03)^2}{2 \times 100}$$ $$m = \frac{2 \times 10^{-6} \times 16 \times 10^{-6} \times 9 \times 10^{-4}}{200}$$ $$m = \frac{288 \times 10^{-16}}{200} = 1.44 \times 10^{-16} \text{ kg} = 144 \times 10^{-18} \text{ kg}$$

Hence, the mass of the charged particle is $$144 \times 10^{-18}$$ kg. So, the answer is $$144$$.