Problem 7.27. A charge of - 1.7 x 10-’ C is moving west a velocity of 3 x 105 m/

Question

Problem 7.27. A charge of - 1.7 x 10-’ C is moving west a velocity of 3 x 105 m/s. What magnetic field (magnitude and direction) does it produce at a point, P, which is reached by going north 1.2 x 10-2 m and then west by 0.9 x 10-2 m? (See Fig. 7-28.)

Ans. 0.18 T, vertically up

Problem 7.28. An elevator, carrying a charge of 0.2 C, is moving down with a velocity of 4 x 103 m/s. The elevator is 10 m from the bottom and 3 m horizontally from point P in Fig. 7-29. What magnetic field does it produce at point P?

Ans. 2.1 x 1OP5 T, out

Explanation / Answer

7.27)
B = (miuo/ 4pi) * q*v sin thetha / r^2
thetha is angle between v anmd r
Its direction will be q*(vXr)
vXr points down butr since q is negative, B will point up
so,
direction upward

r = sqrt ((1*10^-2)^2 + (0.9*10^-2)^2)
=0.0135 m

angle = atan (0.9*10^-2 /1.2*10^-2) = 37 degree
B = (miuo/ 4pi) * q*v sin thetha / r^2
= 10^-7 * q*(3*10^5)* sin 37 / (0.0135)^2
= 0.18 T
Answer: 0.18 T

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