Question

A 2MeV proton is moving perpendicular to uniform magnetic field of 2.5T. The magnetic force on the proton :-

• A. 8 x 10-12 N
• B. 4 x 10-12 N
• C. 16 x 10-12 N
• D. 2 x 10-12 N

Answer 1

Answer: (A)

8 x 10-12 N

Answer 2

To calculate the magnetic force on the proton, we need to determine its velocity first using its kinetic energy.

1. Convert Kinetic Energy to Joules: Given kinetic energy (KE) = 2 MeV. We know that 1 MeV = $10^6$ eV and 1 eV = $1.6 \times 10^{-19}$ J. So, KE = $2 \times 10^6 \times 1.6 \times 10^{-19}$ J KE = $3.2 \times 10^{-13}$ J

2. Calculate the Velocity of the Proton: The kinetic energy formula is $KE = \frac{1}{2}mv^2$, where 'm' is the mass of the proton and 'v' is its velocity. The mass of a proton ($m$) = $1.67 \times 10^{-27}$ kg. Rearranging the formula to find 'v': $v^2 = \frac{2 \times KE}{m}$ $v^2 = \frac{2 \times 3.2 \times 10^{-13} \text{ J}}{1.67 \times 10^{-27} \text{ kg}}$ $v^2 = \frac{6.4 \times 10^{-13}}{1.67 \times 10^{-27}}$ $v^2 \approx 3.8323 \times 10^{14}$ $v = \sqrt{3.8323 \times 10^{14}}$ $v \approx 1.9576 \times 10^7 \text{ m/s}$

3. Calculate the Magnetic Force: The magnetic force (F) on a charged particle moving in a magnetic field is given by the formula: $F = qvB \sin\theta$ Where:

• $q$ is the charge of the proton = $1.6 \times 10^{-19}$ C
• $v$ is the velocity of the proton $\approx 1.9576 \times 10^7$ m/s
• $B$ is the magnetic field strength = 2.5 T
• $\theta$ is the angle between the velocity vector and the magnetic field. Given that the proton is moving perpendicular to the magnetic field, $\theta = 90^\circ$, so $\sin\theta = \sin 90^\circ = 1$.

Substitute the values into the formula: $F = (1.6 \times 10^{-19} \text{ C}) \times (1.9576 \times 10^7 \text{ m/s}) \times (2.5 \text{ T}) \times 1$ $F = (1.6 \times 2.5 \times 1.9576) \times 10^{(-19+7)}$ $F = (4 \times 1.9576) \times 10^{-12}$ $F = 7.8304 \times 10^{-12}$ N

Rounding this value, it is approximately $8 \times 10^{-12}$ N.