Question

A proton travels with a speed of 3.00 x 10 m/s at an angle of 34.0° with the direction of a magnetic field of 0.400 T in the y direction. What is the magnitude of the magnetic force on the proton? (Your result must be in multiples of 10-14 N and include 1 digit after the decimal point. That means if you get a result of a 92.2 x 10-14 just type 92.2 in the answer box. Maximum of 5% of error is accepted in your answer. Take elementary charge e = 1.6x 10-19 C.)

Answer 1

Press charge =te -19 = 1.6*10 no = 3.00 x 106 mis Speed Magnetic field B=0.400T Angle between en magnetic field B and vectocity v = 34.0° Force 111 Proton (magnetic force) = q (NXB) i By definitionſ FI al v XB I (Taking magnitude) quo sin34° lox Charge NB sino o is the angle between a and B) -.q=e IPI= enB sin 34° 1.6x10-19 x3x108x0,400 sim 34 1.0736 x 10 -13 10,736 x 10 10-14 R? 10.7 Ix 1014 N CS Scanned with CamScanner