Aproton travels at a speed of 2.0 × 106 meters/second. its velocity is at right angles with a magnetic field of strength 5.5 × 10-3 tesla. what is the magnitude of the magnetic force on the proton? a. 1.0 × 10-15 newtons b. 1.8 × 10-15 newtons c. 2.1 × 10-16 newtons d. 5.5 × 10-16 newtons

The answer is B : F = 1.8 X 10 ∧ -15

Given:

Velocity of the proton: 2.0 × 10∧6 m/sec                                      

Magnetic field strength(B): 5.5 × 10∧-3 tesla

Now it is given that velocity and the magnetic field are at 90 degree.

Also Magnetic force F= qvBsin∅                                                       

where F is the magnetic force                                                                    

q is the charge of the proton  which is equal to  1.602x10∧-19 coloumbs

∅ is the angle between the v and B.                                                          

v is the velocity of the proton                                                                

B is the magnetic field                                                              

Substituting the values  we get

F = 1.602 x 10 ∧-19 × 2.0 × 10∧6 × 5.5 × 10∧-3 Sin 90

F= 17.6 x 10 ∧-16  N                                                                                        

F= 1.76 x 10∧-15  N

Rounding off we get  F = 1.8 X 10 ∧ -15 N

PLATO ANSWERS

1.8 × 10^6-15 newtons

90.44 m

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 30 m/s

Angle of projection (θ) = 40°

Acceleration due to gravity (g) = 9.8 m/s²

Range (R) =?

The range of the ball can be obtained as follow:

R = u² Sine 2θ / g

R = 30² Sine (2×40) / 9.8

R = 900 Sine 80 / 9.8

R = 900 × 0.9848 / 9.8

R = 886.32 / 9.8

R = 90.44 m

Therefore, the range of the tennis ball is 90.44 m