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A point pole has a strength of 4? × 10-4 weber. The force in newtons on a point pole of 4? × 1.5 × 10-4 weber placed at a distance of 10 cm from it will be a.) 20 N. b.) 15 N. c.) 7.5 N. d.) 3.75 N.

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Solution: m1 = 4N × 10^-7, m2 = 4N × 1.5 × 10^-4; r = 10 cm. F= (m1 m2)/( 4 pi mue0 r^2) = (4N 10^-7 x 4N 10^^-4)/(4N x 4N x 10^-7(10 x10^-2)^2 =15N
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Kindly re-post the question without question mark symbol.
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Don't want to over express myself, just join and feel the experience.

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A software engineer with more that 12 years of experience and 5 years of teaching experience

I am assuming that the 4? actually means 4? . If so the correct answer is b.) 15N. using F =m1m2/4?µ0r^2 we get 14.99N
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b
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Kindly re-post the question without question mark symbol.
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