Question

A proton travels with a speed of 5.00 x 106 m/s at an angle of 61° with the direction of a magnetic field of magnitude 0.150 T in the positive x-direction. (a) What is the magnitude of the magnetic force on the proton? (b) What is the proton's acceleration? (Give the magnitude.) Need Help? Read

Answer 1

Here we have given Velocity of prolon V = 5:00 x 100 mis Magnetic field B = 0.150 T Angle between VfB is o=61° And we know charge on proton mass of proton q = 1.681019c m=1-67x1027 kg a) magnetic force on charged particle is given by Fmax = q (L x B) = quo sino â Where ñ gives direction of force magnitude of force is Finey = quosino so a. Emas = 16X10'965 R106x0.15* (Sin 61°) = 112 x 10-13 X0-8746 N nay = 1.05 * 1013 N ² mă 6) according to Newtons and law Prati T i a = F where a is magnitude of ă cacceleralimall and F is magnitude of Ř (Force) 13 a = 105 x 1067X1027 La = 0628 1014 m/s2 [a= 6.28x1013 m/s2