What is the angle between the proton's velocity and the field?

Question

A proton moving at 3.00*10^6 m/s through a magnetic field of 1.40 T experiences a magnetic force of magnitude 8.20*10^-13 N.

Answer 1

F=q V x B
where
F - The force produced, measured in newtons
q - Electric charge that the magnetic field is acting on, measured in coulombs
V - Velocity of the electric charge , measured in metres per second
B - Strength of the magnetic field


We can also write the vector cross product as
F=q |V| |B| sin(A)
Since protron’s charge = to that of electron
q=1.602 e-19 C
A= arcSin( F/[ q |V| |B|])
A=arcSin(8.20*10^-13 N /[ 1.602 e-19 C x 3.00*10^6 m/s x 1.40 T])
A= arcSin(8.20*e-13 N /[ 6.72 840 e - 13]
A=arcSin(1.2) not on any life time the argument must be less or equal to 1. There is a mistake in the problem statement.

Please let me know how things will work themselves out.

Answer 2

stress on a shifting fee lower than a consistent magnetic field(B) is given through the relation F = q(V*B) Or F= qVBsin(theta) the position F=stress on fee(q) shifting with velocity(v) In case if fee strikes parlal to field F= 0 And if perspective between them is ninety then stress will be optimal and the fee will pass in this style of way that this stress will provide mandatory centrishtal stress i.e. In a round course And if perspective lie b/w 0 & ninety fee will pass in a lengthy spring like course. And in all above description magnetic stress in trouble-free words variations the path of the shifting fee no longer the cost. And hence it does no longer effact the kinetic potential and hence it is speedic potential and hence it is velocity