A proton travels through uniform magnetic and electric fields. The magnetic fiel

Question

A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 2.57 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2920 m/s. At that instant, what is the magnitude of the net force acting on the proton if the electric field is (a) in the positive z direction and has a magnitude of 4.02 V/m, (b) in the negative z direction and has a magnitude of 4.02 V/m, and (c) in the positive x direction and has a magnitude of 4.02 V/m?

Explanation / Answer

F(net) = qE +q[ v cross B]
----------------------------
q=1.6*10^-19
B = 2.57*10^-3 (- i) Tesla >>>>> vector form with unit vectors
v = 2920 (+ j) m/s

a)
E = 4.02 (+ k) V/m
-----------------------
F(net) = q[4.02 (k) + 2920 (j) cross 2.57*10^-3 ( - i)]
F(net) = q[4.02 (k) - 2920*2.57*10^-3 {(j) cross (i)}]
F(net) = q[4.02 (k) + 7.5044 {k}] = 11.5244*q k
=========================
i cross j = k
j cross i = - k
------------------
F(net) = 11.5244*1.6*10^-19 k
|F(net)| = 18.43904*10^-19 N
==================
b) E = 4.02 (- k) V/m
-----------------------
F(net) = q[4.02 (- k) + 2920 (j) cross 2.57*10^-3 ( - i)]
F(net) = q[ - 4.02 (k) + 7.5044 {k}]
=========================
F(net) = 3.4844*1.6*10^-19 k
|F(net)| = 5.57504*10^-19 N
\\\\\\\
c)
E = 4.02 ( i) V/m
-----------------------
F(net) = q[4.02 (i) + 2920 (j) cross 2.57*10^-3 ( - i)]
F(net) = q[4.02 (i) + 7.5044 {k}]
=========================
|F(net)| = q ?[4.02^2 + 7.5044^2] = 8.5133 q
= 8.5133 *1.6*10^-19
= 13.62129*10^-19 N

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