A particle with charge 7.80 μC is moving with velocity v=-3.80×103 m/sj^. the magnetic force on the particle is measured to be F=+7.60×10-3 Ni^-5.20×10-3 Nk^. Calculate the components of the magnetic field you can find from this information.

F=qv×B 7.6×10-3i^-5.2×10-3k^ =7.8×10-63.8×10-3j^Bxi^+Byj^+Bzk^ ∴7.8×10-63.8×10-3Bx =-5.2×10-3 ∴Bx=-0.175 TSimilarly,7.8×10-6-3.8×10-3Bz =7.6×10-3 Bz=-0.256 T

A particle with charge 7.80 μC is moving with velocity v=-3.80×103 m/sj^. the magnetic force on the particle is measured to be F=+7.60×10-3 Ni^-5.20×10-3 Nk^. Calculate the components of the magnetic field you can find from this information.

F=qv×B 7.6×10-3i^-5.2×10-3k^ =7.8×10-63.8×10-3j^Bxi^+Byj^+Bzk^ ∴7.8×10-63.8×10-3Bx =-5.2×10-3 ∴Bx=-0.175 TSimilarly,7.8×10-6-3.8×10-3Bz =7.6×10-3 Bz=-0.256 T

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A particle with charge $7.80 μ C$ is moving with velocity $v = - (3.80 \times 10^{3} m / s) \hat{j}$ . the magnetic force on the particle is measured to be $F = + (7.60 \times 10^{- 3} N) \hat{i} - (5.20 \times 10^{- 3} N) \hat{k}$ . Calculate the components of the magnetic field you can find from this information.

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$B_{x} = 0.175 T$ and $B_{z} = - 0.256 T$

$B_{x} = - 0.175 T$ and $B_{z} = - 0.256 T$

$B_{x} = - 0.175 T$ and $B_{z} = 0.256 T$

$B_{x} = - 0.256 T$ and $B_{z} = 0.256 T$

answer is A.

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Detailed Solution

$F = q (v \times B) [(7.6 \times 10^{- 3}) \hat{i} - (5.2 \times 10^{- 3}) \hat{k}] = (7.8 \times 10^{- 6}) [(3.8 \times 10^{- 3}) \hat{j} (B_{x} \hat{i} + B_{y} \hat{j} + B_{z} \hat{k})] ∴ (7.8 \times 10^{- 6}) (3.8 \times 10^{- 3}) (B_{x}) = (- 5.2 \times 10^{- 3}) ∴ B_{x} = - 0.175 T$