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Question

A charge particle of charge q = 2 C has velocity v=100ms1. When it passes point A adn has velocity in the direction shown. The strength of magnetic field at the point B due to this moving charge is (r = 2m)
1280769_24689dcbae234ff8b787bf63185eaebe.png

A
2.5μT
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B
5.0μT
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C
2.0μT
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D
None of these
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Solution

The correct option is A 2.5μT
Given :charge(q)=2C,v=100m/s, r=2m,θ=30v=100ms−1
Solution: We know the formula of magnetic field on moving charge is :
B=μ0M4πr3
And as we know the know the formula of magnetic moment in terms of q &v
M=q(v×r)
B=μ0q(vrsin300)4πr3=107×2×100×14×2=2.5μT
Hence the correct answer is A

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