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An electron, accelerated before by a potential difference U=300V is moving parallel...
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The electon accelerated by a potential of U=300 V has qU of kinetic energy.
So 300q = (1/2)mv^2 ---> the electron is traveling v = 0.10x10^8 m/s (or 3% of the speed of light).
The magnetic field induced by current in the wire is : B = uI/2PIr. So B = (4PIx10-7 H/m * 5A)/(2PI * 0.004 m) = 250x10-6 Wb/m^2 = 250 uT.
The force on a charged particle moving through an electrical and magnetic field is: F = q(E + v x B). Since the motion of the electron is parallel to the wire, the cross product reduces to vB. The E field outside of the conductor is zero.
So, F = -q0.1x10^8 m/s * 250x10-6 T = -2500 N
Posted by kjcdb8er on July 1, 2009 at 6:54 AM (Answer #1)
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