Question

One component of a magnetic field has a magnitude of 0.062 T and points along the x axis while the other component has a magnitude of 0.036 T and points along the y axis A particle carrying a charge of 1.50 10 5 C is moving along the z axis at a speed of 5.80 103 m s.
a)What is the magnitude of the net magnetic force that acts on the particle?b)What is the angle that the net force makes with respect to the +x axis?

Answer 1

Answer:

A) 62.37 x 10^(6) N

B) 59.86°

Explanation:

A) The magnitude of the force F1 due to the 0.062T magnetic field is;

F1 = |qo|(v)(Bx) Sin 90°

F1 = 1.50 x 10^(5)C x 5.8 x 10³m/s x 0.062T = 53.94 x 10^(6) N

Also, the magnitude of the force F2 due to the 0.036T magnetic field is;

F2 = |qo|(v)(By) Sin 90°

F2 = 1.50 x 10^(5)C x 5.8 x 10³m/s x 0.036T = 31.32 x 10^(6) N

Now, using Fleming's right hand rule, we will discover that the net force makes an angle with the x-axis. Let's call the angle θ. Since the forces will be at right angles to each other, then using Pythagoras theorem will get the resultant met force.

Thus;

F =√((F1)² + (F2)²) = √((53.94 x 10^(6))² + (31.32 x 10^(6))²)

= 62.37 x 10^(6) N

B) Since the forces are at right angles to each other, they are opposite and adjacent and since we know that tangent = opp/adj;

tan θ = F1/F2

tan θ = (53.94 x 10^(6)) / 31.32 x 10^(6) = 1.7222

Thus,θ = tan^(-1) 1.7222

θ = 59.86°