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Please I need help with this Physics question! An alpha particle, entering a uniform external magnetic field?

• An alpha particle, entering a uniform external magnetic field of 1.75 T, begins to move in a circular path of radius 0.0720 m. The magnitude of the electric field required to allow the alpha particle to pass straight through is: I just need steps!

• Answer:

  If the alpha particle is to pass straight through, there must be no net force on it (Newton's First Law). Therefore, the magnetic and electric forces must be equal and opposite. If the magnitudes are equal F_B = F_E then qvB sinφ = qE So E = vB sinφ If the alpha particle moved in a circle before the electric field was introduced, then v and B must have been perpendicular, making sinφ = 1, so E = vB Ah, then we need v, don't we? Apply Newton's Second Law to the alpha particle when it was moving in a circle. Sketch a Free Body Diagram. There is only one force on the alpha particle, the magnetic force which points toward the center of the circular path. Choose your axis to point in that same direction, because that's the known direction of the acceleration. I like to call that the "c" axis, as it points toward the center of the circle. Write out Newton's Second Law for that direction: Σ F_c = F_B = m a_c We know the centripetal acceleration is a_c = v²/R So qvB sinφ = mv²/R Solve for v, remembering that we've already decided that sinφ = 1 v = qBR / m Substitute this into the relationship we found between E and B E = vB ⇒ E = ( qBR / m ) B = qB²R / m B and R were given in the problem, and since the particle is an alpha particle, q and m can be looked up. Be sure to convert q to C and m to kg if you want your electric field in V/m or N/C.