Question

A proton is projected with a speed of $2 \times 10^{6}(\mathrm{~m} / \mathrm{s})\( at an angle of \)60^{\circ}\( to the \)\mathrm{X}$ -axis. If a uniform mag. field of \(0.104\) tesla is applied along \(\mathrm{Y}\) -axis, the path of proton is (a) A circle of \(\mathrm{r}=0.2 \mathrm{~m}\) and time period $\pi \times 10^{-7} \mathrm{sec}$ (b) A circle of \(\mathrm{r}=0.1 \mathrm{~m}\) and time period $2 \pi \times 10^{-7} \mathrm{sec}$ (c) A helix of \(r=0.1 \mathrm{~m}\) and time period $2 \pi \times 10^{-7} \mathrm{sec}$ (d) A helix of \(\mathrm{r}=0.2 \mathrm{~m}\) and time period $4 \pi \times 10^{-7} \mathrm{sec}$

Short answer

The path of the proton is a helix with a radius of \(0.1 \, \text{m}\) and a time period of \(2 \pi \times 10^{-7} \, \text{s}\).

Step-by-step solution

Calculate the force due to the magnetic field

To calculate the magnetic force acting on the proton, let's use the Lorentz force formula: \[F = qvB\sin\theta\] where - \(F\) is the magnetic force, - \(q\) is the charge of the proton, - \(v\) is the speed of the proton, - \(B\) is the magnetic field strength, and - \(\theta\) is the angle between the velocity vector and the magnetic field. Since we know that the proton has a velocity of \(2\times10^6 \frac{\text{m}}{\text{s}}\), we can calculate the magnetic force acting on it. For a proton, the charge \(q = 1.6\times10^{-19} \, \text{C}\). We are given the magnetic field strength, \(B = 0.104 \, \text{T}\). The angle between the proton's velocity and the magnetic field is given as \(60^\circ\). Therefore, the magnetic force on the proton would be: \[F = (1.6\times10^{-19} \, \text{C})(2\times10^6 \frac{\text{m}}{\text{s}})(0.104 \, \text{T})\sin{60^\circ}\]

Calculate the centripetal force for circular motion

Since the magnetic force acting on the proton is responsible for its circular motion, we can equate it to the centripetal force on the proton. The centripetal force is given by: \[F_c = \frac{mv^2}{r}\] Here, - \(F_c\) is the centripetal force, - \(m\) is the mass of the proton, - \(v\) is the speed of the proton, and - \(r\) is the radius of the circle. For a proton, the mass \(m = 1.67\times10^{-27} \, \text{kg}\), and we already calculated the magnetic force on the proton. Let's set the magnetic force equal to the centripetal force and solve for the radius of the circle (\(r\)): \[F_c = F \Rightarrow \frac{1.67\times10^{-27} \, \text{kg}(2\times10^6 \frac{\text{m}}{\text{s}})^2}{r} = (1.6\times10^{-19} \, \text{C})(2\times10^6 \frac{\text{m}}{\text{s}})(0.104 \, \text{T})\sin{60^\circ}\]

Find the radius of the circular path

Solving the equation in Step 2 for \(r\), we find the radius of the circle: \[r = \frac{1.67\times10^{-27} \, \text{kg}(2\times10^6 \frac{\text{m}}{\text{s}})^2}{(1.6\times10^{-19} \, \text{C})(2\times10^6 \frac{\text{m}}{\text{s}})(0.104 \, \text{T})\sin{60^\circ}}\] \[r ≈ 0.1 \, \text{m}\]

Calculate the time period of circular motion

Since the proton is moving in a circle, we need to find the time period of its motion. Using the formula for the time period of circular motion, we get: \[T = \frac{2 \pi r}{v}\] Where \(T\) is the time period, and we already know the radius of the circle (\(r\)) and the speed of the proton (\(v\)). Now we can substitute the values to find the time period: \[T = \frac{2 \pi (0.1 \, \text{m})}{2\times10^6 \frac{\text{m}}{\text{s}}}\] \[T ≈ 2 \pi \times 10^{-7} \, \text{s}\] Now that we have the radius and time period, we can see that the correct answer is (c) A helix of \(r = 0.1 \, \text{m}\) and a time period of \(2 \pi \times 10^{-7} \, \text{s}\).