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Chapter 27 Problem 51 † Given ~ = bt k̂ B b = 2.1 T /ms t = 0.40 µs = 4.0 × 10−7 s x = 5.0 cm = 5.0 × 10−2 m y=0 z=0 ~v = 4.8 × 106 ĵ m/s Solution Find the net electromagnetic force on the proton. First convert the value of b into standard units. 2.1 T 1000 ms = 2100 T /s b= ms 1s At t = 0.40 µ s the magnetic field is ~ = (2100 k̂ T /s)(4.0 × 10−7 s) = 8.4 × 10−4 k̂ T B The magnetic force on the proton is then ~ F~m = q~v × B F~m = (1.6 × 10−19 C)(4.8 × 106 ĵ m/s) × (8.4 × 10−4 k̂ T ) = 6.5 × 10−16 î N Because the magnetic field is changing, an electric field is generated. I ~ r = − dΦB Ed~ dt The magnetic flux for a circle of radius, r is ~·B ~ = AB cos θ = πr2 B cos θ = (πr2 )bt cos(0) = πr2 bt ΦB = A Therefore, dΦB /dt is dΦB d((πr2 bt) = = πr2 b dt dt When travelling around a circle centerred on the middle of the solenoid, the Electric field is constant and the line integral becomes I I ~ Ed~r = E dr = E2πr Setting these equal to each other and solving for the electric field gives E2πr = −πr2 b E=− † πr2 b rb (5.0 × 10−2 m)(2100 T m2 /s) =− =− = −52.5 N/C 2πr 2 2 Problem from Essential University Physics, Wolfson Since the magnetic field is initially upward and increasing, the induced electric field is oriented so it would cause a clockwise current to flow and thus generate a downward magnetic field (by Lenz’s law). This opposition to change is why the negative sign shows up in the previous calculation. Therefore, at the location of the proton (x = 5.0 cm y = 0)the electric field vector is ~ = −52.5 ĵ N/C E The electric force is then ~ = (1.6 × 10−19 C)(−52.5 ĵ N/C) = −8.4 × 10−18 ĵ N F~e = q E The total force on the proton is F~ = {−8.4 × 10−18 ĵ + 6.5 × 10−16 î} N