A uniform magnetic field B exists in the +ve x direction. A proton (q =+e) shoots through the field in the +ve y direction with the speed v. Calculate v.

Explanation

When a charged particle, such as a proton, moves through a magnetic field, it experiences a force known as the magnetic Lorentz force. The magnitude of this force can be calculated using the equation:F = q * v * B * sin(theta),where F is the magnitude of the magnetic force, q is the charge of the particle, v is its velocity, B is the magnitude of the magnetic field, and theta is the angle between the velocity vector and the magnetic field vector.In this case, the proton is moving in the +ve y direction, perpendicular to the magnetic field in the +ve x direction. Therefore, the angle between the velocity vector and the magnetic field vector is 90 degrees, and sin(theta) = 1.The magnetic force experienced by the proton is equal to the centripetal force required to keep it moving in a curved path. The centripetal force is given by the equation:F_c = (m * v^2) / r,where m is the mass of the proton, v is its velocity, and r is the radius of the curved path.Since the magnetic force is equal to the centripetal force, we can equate the two equations:q * v * B = (m * v^2) / r.Simplifying the equation, we find:v = (q * B * r) / m.The velocity of the proton is directly proportional to the magnetic field strength (B), the radius of the curved path (r), and the charge of the proton (q), and inversely proportional to the mass of the proton (m).Therefore, option B, equal to the force on the proton, is the correct answer. The velocity of the proton is determined by the magnetic field, the radius of the curved path, the charge of the proton, and the mass of the proton.