The magnitude of the magnetic field that will cause the proton to follow the semicircular path from A to B is 0.294 T and the direction should be inwards.
According to the attached figure, the proton follows a semicircular path from point A to B. The velocity of proton is Vo = 1.41 x 10^6 m/s. The figure shows that the diameter of the circular path is 10cm, hence the radius of the path is 5 cm or 0.05m. The magnetic force experienced by the proton is balanced by the centripetal force acting on it. The magnetic force is given by qvB and the centripetal force is given by mv^2/r. Hence,
(mv^2)/r=qvB
Where m is the mass of the particle, v is velocity of the particle, r is the radius, q is charge of moving particle and B is the magnitude of magnetic field. Thus
B=mv/rq
Substituting the values,
B=(1.67 x 10^(-27)*1.41 x 10^6)/(0.05*1.602 x 10^(-19) )
B = 0.294 T
Hence, the magnitude of magnetic field is 0.294 T.
Magnetic force at point A should be towards right. Hence, from Fleming's left hand rule magnetic field should be inwards.
Note: The question is incomplete. The complete question probably is: A proton at point A in figure has a speed Vo = 1.41 x 10^6 m/s. Find the magnitude and direction of the magnetic field that will cause the proton to follow the semicircular path from A to B.
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