Question

An antiproton (same properties as a proton except that q=?e) is moving in the combined electric and magnetic fields of the figure.(Figure 1) Assume that B = 2.5 T and E = 1000 V/m .

What is the magnitude of the antiproton's acceleration at this instant?

Express your answer to two significant figures and include the appropriate units.
ㄨㄨㄨㄨㄨ 500 m/s

What is the direction of the antiproton's acceleration at this instant?

Answer 1

the force on the antiproton due to the magnetic field

$F_B = q(V \times B) = (1.6\times 10^{-19})(500 \times 2.5) = 2 \times 10^{-16} N$

The force is upward

Force due ot electric field

$F_E = qE = (1.6\times 10^{-19})(1000) = 1.6 \times 10^{-16} N$

The force is downward. so the net force is

$a = \frac{\Delta F}{m} =\frac{0.4 \times 10^{-16} }{1.67 \times 10^{-27}} = 2.4 \times 10^{10} m/s^2$

net force is upward so is the acceleration.