Question

18) A proton travels with a speed (v) of 5m / s in a 2T magnetic field (B). The angle between v
and B is 22nd. What is the magnitude of the magnetic force acting on the proton?
(a) 14.8x10-19N (b) 6x10-19N (c) 0N (d) 16x10-19N
19) An electron experiences a magnetic force FB = (i 2.4 - j 1.8) x10-13N when it passes through a field
magnetic B = k 0.7T. The magnitude of the electron velocity is:
(a) 1.6x106m / s (b) 2.7x106m / s (c) 2.1x106m / s (d) 4.3x106m / s

Answer 1

18.

Magnetic force on a moving charge in magnetic field is given by:

F = q*(VxB) = q*V*B*sin

Using given values:

F = 1.6*10^-19*5*2*sin 22 deg

F = 6.0*10^-19 N

Correct option is B.

19.

Again using above equation in vector form:

F = q*(VxB)

F = (2.4 i - 1.8 j)*10^-13 N

q = charge on electron = -1.6*10^-19 C

V = (Vx i + Vy j + Vz k)

B = (0.7 k) T

q*VxB = -1.6*10^-19*(Vx i + Vy j + Vz k)x(0.7 k)

q*VxB = -1.6*10^-19*0.7*Vx* (ixk) - 1.6*10^-19*0.7*Vy*(jxk) - 1.6*10^-19*Vz*0.7* (kxk)

ixk = -j, jxk = i & kxk = 0, So

q*VxB = -1.6*10^-19*0.7*Vx* (-j) - 1.6*10^-19*0.7*Vy*(i) + 0

Now compare both force values:

(2.4 i - 1.8 j)*10^-13 = -1.6*10^-19*0.7*Vy i + 1.6*10^-19*0.7*Vx j

2.4*10^-13 = -1.6*10^-19*0.7*Vy

Vy = (2.4*10^-13)/(-1.6*10^-19*0.7) = -2.14*10^6 m/s

-1.8*10^-13 = 1.6*10^-19*0.7*Vx

Vx = (-1.8*10^-13)/(1.6*10^-19*0.7) = -1.61*10^6 m/s

So

V = -1.61*10^6 i - 2.14*10^6 j

|V| = sqrt ((-1.61*10^6)^2 + (-2.14*10^6)^2)

|V| = 2.7*10^6 m/s

Correct option is B.