Here's how to approach this problem:
1. Calculate the Kinetic Energy
* The potential difference accelerates the particle, giving it kinetic energy. The relationship is:
* ΔKE = qΔV
* Where:
* ΔKE is the change in kinetic energy
* q is the charge of the particle
* ΔV is the potential difference
* Calculate ΔKE:
* ΔKE = (3.20 x 10^-19 C)(2.45 x 10^6 V) = 7.84 x 10^-13 J
2. Calculate the Velocity
* The kinetic energy is related to the particle's velocity:
* KE = (1/2)mv^2
* Where:
* KE is the kinetic energy (which is equal to ΔKE since it started at rest)
* m is the mass of the particle
* v is the velocity of the particle
* Solve for v:
* v = √(2KE/m) = √(2 * 7.84 x 10^-13 J / 6.64 x 10^-27 kg) ≈ 1.54 x 10^7 m/s
3. Determine the Force and Motion in the Magnetic Field
* A charged particle moving in a magnetic field experiences a force given by:
* F = qvB sin θ
* Where:
* F is the magnetic force
* q is the charge of the particle
* v is the velocity of the particle
* B is the magnetic field strength
* θ is the angle between the velocity and the magnetic field
* Since the problem doesn't specify the angle, we'll assume the particle enters the magnetic field perpendicularly (θ = 90°). This means sin θ = 1.
* Calculate the force:
* F = (3.20 x 10^-19 C)(1.54 x 10^7 m/s)(1.60 T)(1) ≈ 7.94 x 10^-12 N
* The motion in the magnetic field: The force on the particle is perpendicular to its velocity, causing it to move in a circular path. The radius of this path (the radius of curvature) is given by:
* r = mv / (qB)
* Calculate the radius of the circular path:
* r = (6.64 x 10^-27 kg)(1.54 x 10^7 m/s) / (3.20 x 10^-19 C)(1.60 T) ≈ 0.201 m
Summary
The particle, accelerated by the potential difference, enters the magnetic field with a velocity of approximately 1.54 x 10^7 m/s. The magnetic field exerts a force on the particle, causing it to move in a circular path with a radius of about 0.201 meters.
