Physics 1120 Prof James Dove CAPA Set 5 Solutions 5 1 The equation for cyclotron motion is r mv qB where v is the component of the velocity perpendicular to the magnetic field which in this case is along the axis of the solenoid Thus v v sin 5 2 The pitch is d vpar P where P is the period of the cyclotron orbit P 2 r v 2 m qB vpar is the component of the velocity along the magnetic field so vpar v cos 5 4 The current is given by I Q t Then the magnetic field strength is B 0 I 2 d 5 5 The magnetic field in the solenoid is given by B 0 N I L where N is the number of turns and L is the length of the solenoid Since the field is given you can find the current Then use V IR to determine the resistance the voltage is given 5 6 Need 0 I1 2 r1 0 I2 2 r2 where r1 3a and r2 a 5 7 At location P2 the magnetic fields due to each wire actually add together so the magnitude is B 0 2 a I1 I2 5 8 The net force on the square loop is the sum of the magnetic force on the nearest leg of the square and that of the furthest leg of the square the force due to the two legs perpendicular to the long wire cancel out Since the current in the far leg is in the opposite direction as that in the near leg of the square the net force is F Ib L B r1 B r2 where B r1 0 Ia 2 r1 and B r2 0 Ia 2 r2 where r1 1 m and r2 0 35 m and L 0 25 m 5 9 This is the magnitude of the magnetic field due to a long straight wire 0 I 2 d PLUS the magnetic field due to the loop 0 I 2r so 0 I 1 1 Btot 2r 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 0 I 4R and for each of the semi infinite long wires is 0 I 4 R so the net magnetic field is 0 I 2 B 1 4R 5 12 Using Ampere s law H 0 Ienc so dr B 2 rB 0 IN so B 0 IN 2 r Using Ampere s Law 2 rB 0 Ienc 0 JA 0 b2 I r2 a2 a2 J 0 for r a Thus B 0 I r2 a2 2 r b2 a2 5 14 Here we need to use the vector nature of the magnetic fields as the field due to each wire is not parallel or anti parallel to each other The magnitude of each field is 0 I 2 r where r p R2 d2 4 Now the magnetic field due to the bottom wire will be directed downward and somewhat to the right perpendicular to the line going from the wire to P and the magnetic field due to the top wire is directed upward and to the right So the sum of these two vectors will be to the right B Bx 2 0 I d 2 0 I d 2 0 I sin 2 r r r R2 d2 4 3 2 5 15 The magnetic field is the superposition of two semi circles with different radii the horizontal slice doesn t contribute since I is in the same direction as r there The fields are in opposite directions so we subtract the magnitudes For each semicircle Bi 0 I 4ri so B 0 I 0 I 1 R 1 2R 4 8R
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