Correct question: A uniform magnetic field of 0.25 T points vertically downward. A proton enters the field with a horizontal velocity of 4.0×10⁻² m/s north. What are the magnitude and direction of the instantaneous force exerted on the proton as it enters the magnetic field
Answer:
1.602×10⁻²¹ N west
Step-by-step explanation:
The force on a charge moving in a magnetic field is given as,
F = Bqv.................. Equation 1
Where F = force on the proton, B = magnetic Field, q = charge of the proton, v = velocity of the proton.
Given: B = 0.25 T, v = 4.0×10⁻² m/s, q = 1.602 x 10⁻¹⁹ C
Substitute into equation 2
F = 0.25(4.0×10⁻²)(1.602 x 10⁻¹⁹)
F = 1.602×10⁻²¹ N
Hence the magnitude and direction of the instantaneous force = 1.602×10⁻²¹ N west