Hwk6Solns - Assignment 6 Solutions Chpt 28 6. The magnetic...

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Assignment 6 Solutions Chpt 28 6. The magnetic force on the proton is Fq vB = × r r r where q = + e . Using Eq. 3-30 this becomes (4 × 10 17 )i ^ + (2 × 10 17 )j ^ = e [(0.03 v y + 40)i ^ + (20 – 0.03 v x ) j ^ – (0.02 v x + 0.01 v y )k ^ ] with SI units understood. Equating corresponding components, we find (a) v x = 3.5 ×1 0 3 m/s, and (b) v y = 7.0 ×1 0 3 m/s. 11. Since the total force given by ) ( B v E e F r r r r × + = vanishes, the electric field r E must be perpendicular to both the particle velocity r v and the magnetic field r B . The magnetic field is perpendicular to the velocity, so r r v B × has magnitude vB and the magnitude of the electric field is given by E = vB . Since the particle has charge e and is accelerated through a potential difference V , and 2 /2 mv eV = 2 ve V = . m Thus, () ( )( ) 19 3 5 27 2 1.60 10 C 10 10 V 2 1.2 T 6.8 10 V m. 9.99 10 kg eV EB m ×× == = × × 15. (a) We seek the electrostatic field established by the separation of charges (brought on by the magnetic force). With Eq. 28-10, we define the magnitude of the electric field as ( )( ) | | | | 20.0 m/s 0.030 T 0.600 V/m Ev B = rr . Its direction may be inferred from Figure 28-8; its direction is opposite to that defined by r r v B × . In summary, ˆ (0.600V m)k E =− r which insures that ) ( B v E e F r r r r × + = vanishes.
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Hwk6Solns - Assignment 6 Solutions Chpt 28 6. The magnetic...

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