An electron moves with speed 2×105ms−1 along the positive x-direction in the presence of a magnetic induction B^=i^+4j^−3k^ tesla. The magnitude of the force experienced by the electron in newton is (charge on the electron=1.6×10−19C )

F=q(V→×B→) =1.6×10−19|ijk20014−3|×105 =1.6×10−14[−j(−6)+k(8)] =1.6×10−14×36+64=16×10−14 =1.6×10−13N

An electron moves with speed 2×105ms−1 along the positive x-direction in the presence of a magnetic induction B^=i^+4j^−3k^ tesla. The magnitude of the force experienced by the electron in newton is (charge on the electron=1.6×10−19C )

F=q(V→×B→) =1.6×10−19|ijk20014−3|×105 =1.6×10−14[−j(−6)+k(8)] =1.6×10−14×36+64=16×10−14 =1.6×10−13N

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An electron moves with speed $2 \times 10^{5} {ms}^{- 1}$ along the positive x-direction in the presence of a magnetic induction $\hat{B} = \hat{i} + 4 \hat{j} - 3 \hat{k}$ tesla. The magnitude of the force experienced by the electron in newton is (charge on the electron $= 1.6 \times 10^{- 19} C$ )

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1 .6 \times 10^{- 13}

1.72 \times 10^{- 13}

1 .18 \times 10^{- 13}

1 .28 \times 10^{- 13}

answer is C.

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Detailed Solution

F = q (\vec{V} \times \vec{B})

$= 1.6 \times 10^{- 19} | \begin{matrix} i & j & k \\ 2 & 0 & 0 \\ 1 & 4 & - 3 \end{matrix} | \times 10^{5}$

$= 1.6 \times 10^{- 14} [- j (- 6) + k (8)]$

$= 1.6 \times 10^{- 14} \times \sqrt{36 + 64} = 16 \times 10^{- 14}$

$= 1.6 \times 10^{- 13} N$

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An electron moves with speed 2×105ms−1 along the positive x-direction in the presence of a magnetic induction B^=i^+4j^−3k^ tesla. The magnitude of the force experienced by the electron in newton is (charge on the electron=1.6×10−19C )

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