A charge q=−4μC has an instantanious velocity v→=2i^−3j^+k^×106m/s in a uniform magnetic fieldB¯=2i^+5j^−3k^×10−2T.The force on the charge is

F¯=qv¯×B¯=−4×10−6i^j^k^2−3125−3106×10−2=−4×10−69−5i^−6−2j^+10+6k^106×10−2=−44i^+8j^+16k^×10−2=−0.16i^−0.32j^−0.64k^N

A charge q=−4μC has an instantanious velocity v→=2i^−3j^+k^×106m/s in a uniform magnetic fieldB¯=2i^+5j^−3k^×10−2T.The force on the charge is

F¯=qv¯×B¯=−4×10−6i^j^k^2−3125−3106×10−2=−4×10−69−5i^−6−2j^+10+6k^106×10−2=−44i^+8j^+16k^×10−2=−0.16i^−0.32j^−0.64k^N

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A charge $q = - 4 μ C$ has an instantanious velocity $\vec{v} = (2 \hat{i} - 3 \hat{j} + \hat{k}) \times 10^{6} m/s$ in a uniform magnetic field $\bar{B} = (2 \hat{i} + 5 \hat{j} - 3 \hat{k}) \times 10^{- 2} T$ .The force on the charge is

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$(- 0.16 \hat{i} - 0.32 \hat{j} - 0.64 \hat{k}) N$

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$(- 0.16 \hat{i} + 0.32 \hat{j} + 0.64 \hat{k}) N$

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Detailed Solution

$\bar{F} = q (\bar{v} \times \bar{B})$

$= - 4 \times 10^{- 6} |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & - 3 & 1 \\ 2 & 5 & - 3 \end{matrix}| 10^{6} \times 10^{- 2}$ $= - 4 \times 10^{- 6} ((9 - 5) \hat{i} - (6 - 2) \hat{j} + (10 + 6) \hat{k}) 10^{6} \times 10^{- 2}$ $= - 4 (4 \hat{i} + 8 \hat{j} + 16 \hat{k}) \times 10^{- 2}$ $= (- 0.16 \hat{i} - 0.32 \hat{j} - 0.64 \hat{k}) N$