An electron with kinetic energy2.1 keV moves horizontally eastward into aregion

Question

An electron with kinetic energy2.1 keV moves horizontally eastward into aregion of space in which there is a downward-directed electricfield of magnitude 15.0 kV/m. What is the magnitudeof the (smallest magnetic field that will cause the electron tocontinue to move horizontally? Ignore the gravitational force,which is rather small.
Towards which directiondoes this magnetic field point? (north, south, east, west, updown). Explain.
Is it possible for a protonto pass through this combination of fields undeflected? If so, whatis its momentum? (kg*m/s)
Towards which directiondoes this magnetic field point? (north, south, east, west, updown). Explain.
Is it possible for a protonto pass through this combination of fields undeflected? If so, whatis its momentum? (kg*m/s)

Explanation / Answer

The electric force on the electron is upwards F = qE If B is north then v x B is upwards andthe magnetic force is downwards because of the negative charge onthe electron. q E = q v B   for the forces to cancel B = E / v 1/2 m v2 = K    where K is thekinetic energy of the electron B = E (m / 2 K) K = 2.1 * 10E3 * 1.6 * 10E-19 = 3.36 * 10E-16 J B = 1.5 * 10E4 * (9.11 * 10E-31 / ( 2 * 3.36 *10E-16)) = 1.5 * 10E-4 * (91.1 / 6.72) B = 5.52 * 10E-4 T Both forces (electric & magnetic) will be reversed indirection for the proton giving the same directions for thefield and velocity. As found above v = E / B = 1.5 * 10E4 / 5.52 * 10E-4 = 2.72 *10E7 m/s   (about c / 10) P (momentum) = m v = 1.67 * 10E-27 * 2.72 * 10E7 = 4.54 *10E-20 kg m / s

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