91Ó°ÊÓ

Protons have a charge of \(q = 1.60 \times 10^{-19} C\) and a mass of \(m = 1.67 \times 10^{-27} kg\). We need these values to calculate the Lorentz force acting on the proton. Step 2: Determine the Lorentz force on the proton

The Lorentz force on a charged particle can be calculated using the following equation: \(\vec{F} = q(\vec{v} \times \vec{B})\). In this problem, we are given that the magnetic field is along the z-axis, and the proton enters with a velocity in the yz-plane, having components \(4 \hat{\mathbf{j}} + 3 \hat{\mathbf{k}}\). Step 3: Calculate the magnetic field B

Since the proton moves in a helical path, its centripetal force is equal to the Lorentz force. The centripetal force can be calculated as \(F_c = \frac{mv^2}{r}\), where m is the mass of the proton, v is its velocity perpendicular to the magnetic field and r is the radius of the helix. First, we need to find the component of the velocity perpendicular to the magnetic field. In this case, that component is in the y-direction: \(v_y = 4 \times 10^6 m/s\). Hence, we can write the equation as: \(|\vec{F}| = \frac{mv_y^2}{r}= q(v_yB)\) Now, we can solve for B: \(B = \frac{mv_y}{qr}\) Substituting the values, we get \(B = \frac{(1.67 \times 10^{-27})(4 \times 10^6)}{(1.6 \times 10^{-19})(5.0 \times 10^{-2})}\) \(B = 8.38 \times 10^{-3} T\) Step 4: Calculate the time for one trip around the helix

To find the time required for one trip around the helix, we need to use the angular frequency equation: \(\omega = \frac{qB}{m}\). The time for one trip around the helix is the time period T, which is equal to \(T = \frac{2\pi}{\omega}\). Substituting the values for q, B, and m, we get: \(T = \frac{2\pi m}{qB} = \frac{2\pi (1.67 \times 10^{-27})}{(1.6 \times 10^{-19})(8.38 \times 10^{-3})}\) \(T= 6.25 \times 10^{-8} s \) Step 5: Determine the proton's position after a given time

To find the position of the proton after 5.0 × 10^{-7} s, we need to consider the motion in both y and z directions. Since the velocity in the z-direction is constant, we can calculate the displacement in the z-direction as follows: Displacement in z: \(\Delta z = v_z t = (3 \times 10^6)(5.0 \times 10^{-7})\) In the y-direction, the proton moves in a circular path. To find its position at a given time, we need to determine the angle it has traveled, and then use trigonometry to find its position. Angle traveled: \(\Theta = \omega t = \frac{q B t}{m}\) Using this angle, we can determine the positions in y and z plane at the given time: \(y = r \sin(\Theta)\) \(z = \Delta z + r \cos(\Theta) - r\) Substituting the values, we get: \(y = (5.0 \times 10^{-2})\sin\left(\frac{(1.6 \times 10^{-19})(8.38 \times 10^{-3})(5.0 \times 10^{-7})}{1.67 \times 10^{-27}}\right)\newline y \approx 4.2 \times 10^{-2} m\) \(z= 1.5 \times 10^{-1} + (5.0 \times 10^{-2})\cos\left(\frac{(1.6 \times 10^{-19})(8.38 \times 10^{-3})(5.0 \times 10^{-7})}{1.67 \times 10^{-27}}\right) - (5.0 \times 10^{-2}) \newline z \approx 0.125 m\) Therefore, the proton's position at t = 5.0 × 10^{-7} s is approximately 4.2 cm in the y-direction and 12.5 cm in the z-direction.