A proton travels to the east at 3.5*10⁶ m/s as it enters a region of uniform magnetic field. The magnitude of the magnetic field is 0.15 T and its direction is 30° north of east. Find the magnitude of the force on the proton at the moment it enters the magnetic field.
Force=qvB sin theta where theta is the angle between v and B
To find the magnitude of the force on the proton, we can use the formula for the magnetic force on a moving charged particle in a magnetic field:
F = q * v * B * sinθ
Where:
F = Force on the charged particle
q = Charge of the particle
v = Velocity of the particle
B = Magnetic field strength
θ = Angle between the velocity vector and the magnetic field vector
In this case, we have a proton, which has a charge of +1.6 × 10⁻¹⁹ C. The velocity of the proton is given as 3.5 × 10⁶ m/s, and the magnetic field strength is 0.15 T. The angle between the velocity vector and the magnetic field vector is 30°.
Now, plug in the values into the formula:
F = (1.6 × 10⁻¹⁹ C) * (3.5 × 10⁶ m/s) * (0.15 T) * sin(30°)
Before evaluating this expression, remember that sin(30°) = 0.5.
F = (1.6 × 10⁻¹⁹ C) * (3.5 × 10⁶ m/s) * (0.15 T) * 0.5
F = (0.84 × 10⁻¹³ N)
Therefore, the magnitude of the force on the proton at the moment it enters the magnetic field is 0.84 × 10⁻¹³ N.