Question

A proton travels through a region of uniform magnetic field at an angle θ relative to the magnetic field. The magnitude...


A proton travels through a region of uniform magnetic field at an angle θ relative to the magnetic field. The magnitude of the magnetic field in this region is 1.15 T, and the proton's velocity is 500,000 m/s when it experiences an acceleration whose magnitude is 1.57 x10^{13} m/s^2. Calculate the angle θ. Give your answer in degrees.

Answer

Mass of proton (m) = 1.67×10-27 kg Charge of proton (q) = 1.6×10-19 CMagnetic field strength (B) = 1.15 TVelocity of proton (v)= 500000 msAcceleration of proton (a)= 1.57×1013 ms2

 

Using the formula for magnetic force : F= q×v×B×SinθUsing newton's second law of motion : net force = mass×acceleration mass×acceleration =q×v×B×Sinθa = q×v×B×Sinθm1.57 ×1013 = 1.6×10-19×500000×1.15×Sinθ1.67×10-271.57×1013 = 5.508×1013×Sinθ0.285 = Sinθ16.55o = θ(Angle between velocity and magnetic field)

 

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