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

Question

What is the magnitude of the antiproton's acceleration at this instant? What is the direction of the antiproton's acceleration at this instant? Toward the top of the screen or toward the bottom of the screen? An antiproton (same properties as a proton except that q=?e q = - e) is moving in the combined electric and magnetic fields of the figure.(Figure 1) Assume that BB B = 2.5T T and EE E = 1100 v/m What is the magnitude of the antiproton's acceleration at this instant? What is the direction of the antiproton's acceleration at this instant? Toward the top of the screen or toward the bottom of the screen?

Explanation / Answer

Let Fb be magnetic force and Fe be electric force.

Fb = q(v X B)

v = -500 i m/s

B = -2.5 k T

Fb = -1.6*10^-19 (-1250 j) = 2 * 10^-16 N j (j indicates force is towards top)

Fe = q*E = -1.6*10^-19*1100 = -1.76*10^-16 N j

Magnitude of antiproton acceleration = (Fb + Fe )/Mp = (0.24*10^-16)/(1.67*10^-27)

= 1.437 * 10^10 m/s^2

Direction is towards top.

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