A charge particle A of charge q = 2C has velocity V = 100 m/s. When it passes through point A and has velocity in the direction shown. The strength of instantaneous magnetic field at point B (r = 2m) due to this moving charge is 0.5×n μT. Find n.

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answer is 0005.00.

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Detailed Solution

dB→=μ04πIdℓ→×r→r3dB→=μ04πdqdtdℓ→×r→r3=μ04π(dq)dℓ→dt×r→r3=μ04πqu→×r→r3=μ0qvsin⁡θ4πr2=2.5×10−6T

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