A2pC charged particle is moving with a speed of 10 m/s. When would it experience

Question

A2pC charged particle is moving with a speed of 10 m/s. When would it experience a greater magnetic force: Case A) if it moved at an angle of 1.0degree with respect to a 17.0 T magnetic field, or Case B) if it moved at an angle of 89.0degree with respect to a 2.7 T magnetic field? A straight wire of length L carries a current l in the positive z direction in a region where the magnetic field is uniform and specified by B_x = 3B, B_y=-2B, and B_z = B, where B is a constant. What are the x-, y-, and z-components of the total force on the wire? What value of B should be used in a velocity selector (shown below) so that protons with a kinetic energy of 2.0-keV are not deflected to the left or the right, if E is fixed at 80 kV/m?

Explanation / Answer

1)

F1 = q*v*B1*sin(theta1)

= 2*10^-12*10*17*sin(1)

= 5.93*10^-12 N

F2 = q*v*B2*sin(theta2)

= 2*10^-12*10*2.7*sin(89)

= 5.40*10^-11 N

clearly F2 > F1

2)

F = I*(L cross B)

= I*( (L k) cross (3*B i - 2*B j + Bk) )

= I*( 3*B*L j - 2*B*L*(-i) + 0 )

= I*B*L*(2i + 3j)

Fx = 2*I*B*L

Fy = 3*I*B*L

Fz = 0

3) speed of proton, v = sqrt(2*KE/m)

= sqrt(2*2*10^3*1.6*10^-19/(1.67*10^-27))

= 6.19*10^5 m/s

in the equilbrium, FB = Fe

q*v*B = q*E

B = E/v

= 80*10^3/(6.19*10^5)

= 0.129 T

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