Physics 1120Prof. James DoveCAPA Set #5 Solutions• (5.1) The equation for cyclotron motion is r = mv⊥/(qB), where v⊥is the component of the velocityperpendicular to the magnetic field, which in this case is along the axis of the solenoid. Thus v⊥=v sin θ.• (5.2) The pitch is d = vparP , where P is the period of the cyclotron orbit, P = 2πr/v⊥= 2πm/qB.vparis the component of the velocity along the magnetic field, so vpar= v cos θ.• (5.4) The current is given by I = ∆Q/∆t. Then the magnetic field strength is B = µ0I/(2πd).• (5.5) The magnetic field in the solenoid is given by B = µ0NI/L, where N is the number of turns andL is the length of the solenoid. Since the field is given you can find the current. Then use ∆V = IRto determine the resistance (the voltage is given).• (5.6) Need µ0I1/(2πr1) = µ0I2/(2πr2), where r1= 3a and r2= a.• (5.7) At location P2, the magnetic fields due to each wire actually add together, so the magnitude isB = µ0/(2πa)(I1+ I2)• (5.8) The net force on the square loop is the sum of the magnetic force on the nearest leg of the squareand that of the furthest leg of the square (the force due to the two legs perpendicular to the long wirecancel out). Since the current in the far leg is in the opposite direction as that in the near leg of thesquare, the net force isF = IbL(B(r1) − B(r2)),where B(r1) = µ0Ia/(2πr1) and B(r2) = µ0Ia/(2πr2), where r1= .1 m and r2= 0.35 m, and L = 0.25m.• (5.9) This is the magnitude of the magnetic field due to a long straight wire (µ0I/(2πd)) PLUS themagnetic field due to the loop (µ0I/(2r)), soBtot=µ0I2r1π+ 1.• (5.11) This is the magnetic field due to half a circular loop and half that of two infinitely long wires(with opposite currents). The magnetic field due to half of the circular loop is µ0I/4R and for each ofthe semi-infinite long wires is µ0I/(4πR), so the net magnetic field isB =µ0I4R2π+ 1• (5.12) Using Ampere’s law,H~B ·~dr = µ0Ienc, so2πrB = µ0IN,so B = µ0IN/(2πr).• Using Ampere’s Law,2πrB = µ0Ienc= µ0JA = µ0Iπ(b2− a2)(π[r2− a2])(J = 0 for r < a) ThusB =µ0I2πrr2− a2(b2− a2)• (5.14) Here we need to use the vector nature of the magnetic fields, as the field due to each wireis not parallel or anti-parallel to each other. The magnitude of each field is µ0I/(2πr), where r =pR2+ d2/4. Now, the magnetic field due to the bottom wire will be directed downward and somewhatto the right (perpendicular to the line going from the wire to P) and the magnetic field due to the topwire is directed upward and to the right. So the sum of these two vectors will be to the right.B = Bx=2µ0I2πrsin(θ) =µ0Iπrd/2r=µ0Iπd/2(R2+ d2/4)3/2• (5.15) The magnetic field is the superposition of two semi-circles, with different radii (the horizontalslice doesn’t contribute since I is in the same direction as ˆr there. The fields are in opposite directions,so we subtract the magnitudes. For each semicircle, Bi= µ0I/(4ri), soB =µ0I4(1/R − 1/2R)
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