JEE Physics 2023 Question with Solution

Given: charge of particle = $2 \, \mu C$, potential difference = $100 \, \text{V}$, magnetic field = $4 \, \text{mT}$, radius of semicircle = $3 \, \text{cm}$.

Find: mass of the particle in the form $\_\_\_\_\_ \times 10^{-18} \, \text{kg}$.

A charged particle accelerated through potential difference $V$ gains kinetic energy:

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

Also, in a magnetic field, the radius of circular path is:

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

So,

$$v = \frac{qBr}{m}$$

Substituting in the energy equation:

$$qV = \frac{1}{2}m\left(\frac{qBr}{m}\right)^2$$ $$qV = \frac{q^2 B^2 r^2}{2m}$$

Hence,

$$m = \frac{q B^2 r^2}{2V}$$

Now substitute:

$$q = 2 \times 10^{-6} \, \text{C}, \quad B = 4 \times 10^{-3} \, \text{T}, \quad r = 3 \times 10^{-2} \, \text{m}, \quad V = 100 \, \text{V}$$ $$m = \frac{\left(2 \times 10^{-6}\right)\left(4 \times 10^{-3}\right)^2\left(3 \times 10^{-2}\right)^2}{2 \times 100}$$ $$m = 144 \times 10^{-18} \, \text{kg}$$

Therefore, the required value is 144.

Given: the particle first gains speed due to electric potential difference and then moves perpendicular to a uniform magnetic field.

Find: the numerical value multiplying $10^{-18} \, \text{kg}$.

From acceleration through potential difference:

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

From circular motion in magnetic field:

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

Rearranging,

$$mv = qBr$$

Squaring both sides,

$$m^2 v^2 = q^2 B^2 r^2$$

Using

$$mv^2 = 2qV$$

we get

$$m(mv^2) = q^2 B^2 r^2$$ $$m(2qV) = q^2 B^2 r^2$$ $$m = \frac{q B^2 r^2}{2V}$$

Substitute numerical values:

$$m = \frac{\left(2 \times 10^{-6}\right)\left(4 \times 10^{-3}\right)^2\left(3 \times 10^{-2}\right)^2}{200}$$ $$= \frac{2 \times 16 \times 9 \times 10^{-6-6-4}}{200}$$ $$= \frac{288 \times 10^{-16}}{200}$$ $$= 1.44 \times 10^{-16} \, \text{kg}$$ $$= 144 \times 10^{-18} \, \text{kg}$$

Therefore, the answer is 144.