Question

A particle with charge 2.10 μC and mass 3.50×10−11 kg is initially traveling in the +y -direction with a speed v0 = 1.60×105 m/s . It then enters a region containing a uniform magnetic field that is directed into, and perpendicular to, the page. The magnitude of the field is 0.410 T . The region extends a distance of 25.0 cm along the initial direction of travel; 75.0 cm from the point of entry into the magnetic field region is a wall. The length of the field-free region is thus 50.0 cm . When the charged particle enters the magnetic field, it follows a curved path whose radius of curvature is R . It then leaves the magnetic field after a time t1 , having been deflected a distance Δx1 . The particle then travels in the field-free region and strikes the wall after undergoing a total deflection Δx . Determinate R, t1, Δx1 e Δx

Answer 1

The radius of the particle's path is 25 centimeters. The time it takes to leave the magnetic field is 1.3 x 10^-11 seconds. The deflection distance is approximately -0.39 centimeters.

Step-by-step explanation:

To determine the radius of the path, we can use the equation for circular motion in a magnetic field: R = (m*v)/(q*B), where R is the radius, m is the mass, v is the velocity, q is the charge, and B is the magnetic field strength. Plugging in the given values, the radius of the path is 0.025 meters or 25 centimeters.

To find the time it takes for the particle to leave the magnetic field, we can use the equation t1 = (m*v)/(q*B). Substituting the given values, the time is calculated to be 1.3 x 10^-11 seconds.

To calculate the deflection distance, we can use the equation Δx1 = (-q*B*t1*v)/(m). Plugging in the given values, the deflection distance is approximately -3.9 x 10^-4 meters or -0.39 centimeters.

The total deflection distance can be found by summing the distance traveled in the field-free region and the deflection distance in the magnetic field. Therefore, Δx = 0.5 meters + (-0.0039 meters) = approximately -0.0034 meters or -0.34 centimeters.