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Q.

Positive point charges g = +8.00 mC and q’ = +3.00 mC are moving relative to an observer at point P, as shown in figure. The distance d is 0.120 m. When the two charges are at the locations shown in the figure, what are the magnitude and direction of the net magnetic field they produce at point P?
(Take v = 4.50 x10⁶ ms^-1 and v’=9.00 x 0⁶ ms^-1)

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a

4.38 x 10^-4 T, into the page

b

4.38 x 10^-4 T, out of the page

c

2.16 x 10^-4 T, into the page

d

2.16 x 10^-4 T, out of the page

answer is A.

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Detailed Solution

$B_{total} = B + B^{'} = \frac{μ_{0}}{4 π} (\frac{q v}{d^{2}} + \frac{q^{'} v^{'}}{d^{2}}) B = \frac{μ_{0}}{4 π} (\frac{(8.0 \times 10^{- 6} C) (4.5 \times 10^{6} {ms}^{- 1})}{(0.120 m)^{2}} + \frac{(3.0 \times 10^{- 6} C) (9.0 \times 10^{6} {ms}^{- 1})}{(0.120 m)^{2}}) B = 4.38 \times 10^{- 4} T, into the page.$

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