The particle experiences a force of 9.2 × 10⁻⁶ N when moving with a speed of 6.2 m/s at an angle of 27 degrees relative to a previously calculated magnetic field.
The force on a charged particle moving through a magnetic field is given by the Lorentz force equation, which is F = qvBsin(θ), where F is the force, q is the charge, v is the velocity, B is the magnetic field strength, and θ is the angle between the velocity vector and the magnetic field vector.
In this case, to find the force when the particle moves at a speed of 6.2 m/s at an angle of 27 degrees with respect to the magnetic field, we use the given information to calculate the magnetic field using its initial condition where the particle was moving perpendicularly (sin(90°) = 1) and then apply the new velocity and angle to find the new force using the calculated magnetic field and the sine of 27 degrees.
Initially, we know that F1 = qv1Bsin(90°), where F1 = 2.1 × 10⁻⁴ N, q = 16 µC, v1 = 28 m/s, and B is what we need to find. We get B = F1/(qv1). Then, using this B, the new force F2 when the particle moves at 6.2 m/s at a 27-degree angle is F2 = qv2Bsin(27°), where v2 = 6.2 m/s. After we calculate B, we can substitute all values into the second equation to get the final answer for F2.
Now, let’s calculate: B = (2.1 × 10⁻⁴ N) / ((16 × 10⁻⁴ C) × 28 m/s) = 0.0046875 T. Using this B, F2 = (16 × 10⁻⁴ C) × 6.2 m/s × 0.0046875 T × sin(27°) = 9.2 × 10⁻⁶ N.
The particle will experience a force of 9.2 × 10⁻⁶ N when it moves at a speed of 6.2 m/s at an angle of 27 degrees to the magnetic field.