Question An electron moves at 2.40 ×10^6 m/s through a region in which there is a magnetic field of unspecified direction and magnitude 7.20 ×10^−2 T.A)What is the largest possible magnitude of the acceleration of the electron due to the magnetic field?B)What is the smallest possible magnitude of the acceleration of the electron due to the magnetic field?C)If the actual acceleration of the electron is one-fourth of the largest magnitude in part AA, what is the angle between the electron velocity and the magnetic field?

Step 1The electron, moving with the given velocity, encounters a magnetic field, at an unspecified direction.Let v be the velocity of the electronv=2.4×106 m/sAnd B be the magnitude of the magnetic fieldB=7.2×10-2 TAs the electron enters the magnetic field, a magnetic force will be exerted on the charged electron, which will force the electron to follow a curved path in the magnetic field. This magnetic force is given by the formulaF=qvBsinθq is the charge on electronθ is the angle between the velocity vector and the direction of the magnetic field Step 2As the electron follows a curved path inside the magnetic field, a centripetal force will also act on the electron, which will result in a centripetal acceleration of the electron.When the electron is at equilibrium, all the forces on the electron cancel each other out. So the magnetic force will thus be equal and opposite to the centripetal force,Fc=macm is the mass of the electronac is its centripetal accelerationAt equilibriummac=qvBsinθac=qmvBsinθNow, sine function has a maximum value 1, and a minimum value 0. So the acceleration of the electron will be minimum when the electron and the magnetic field, both are parallel to each otherIn this case, θ=0sin0=0acmin=0Step 3And maximum possible acceleration of the electron will be attained when the electron and the magnetic field, both remain perpendicular to each other,In this case,θ=90sin90=1ac=qmvBac=1.6×10-199.31×10-31×2.4×106×7.2×10-2ac=1.7186×1011×172800acmax=2.97×1016 m/s2Step 4Given thata=14acmaxa=14×2.97×1016a=7.4244×1015 m/s2Using this value of acceleration,ma=qvBsinθsinθ=maqvBsinθ=9.31×10-31×7.4244×10151.6×10-19×2.4×106×7.2×10-2sinθ=6.912×10-152.7648×10-14sinθ=0.25θ=sin-1(0.25)θ=14.478°Step 5Answers:Maximum possible acceleration of the electron is 2.97 x 1016 m/s2Minimum possible acceleration of the electron is 0Angle between the electron velocity vector and magnetic field is 14.478 ...