Assuming that the MRI scan test involved a magnetic field of 0.1 T, find the maximum and minimum values of the force that this field could exert on a proton moving with a speed of 104ms-4. State the condition under which the force can be minimum.
APPLY THE FORMULA , F(magnitude) = qvBSinQ here, v= 104m/s , q= 1.6 x 10-19 = e. , B = 0.1T
Q= theta = angle b/w the direction of movement of proton and the direction of magnetic field.
THE force is min when the charge moves parallel or antiparallel to the direction of the M field. ie when Q = 0, SINQ=0 then F(MIN) =0N
F(MAX)= qvBSin 90 = qvB =16 x 10-17 N
Q= theta = angle b/w the direction of movement of proton and the direction of magnetic field.
THE force is min when the charge moves parallel or antiparallel to the direction of the M field. ie when Q = 0, SINQ=0 then F(MIN) =0N
F(MAX)= qvBSin 90 = qvB =16 x 10-17 N