A proton of velocity (3i^+2j^)×$$105ms$$−$$1$$ enters a magnitude field $$(2i^+3k^)T$$ . If the specific charge is $$9.6$$×$$107C$$ kg−$$1$$ , the acceleration of the proton in ms−$$2$$ is

Correct option is 3(6i^−9j^−4k^)×9.6×1012

Questions

A proton of velocity $(3 \hat{i} + 2 \hat{j}) \times 10^{5} m s^{- 1}$ enters a magnitude field $(2 \hat{i} + 3 \hat{k}) T$ . If the specific charge is $9.6 \times 10^{7} C k g^{- 1}$ , the acceleration of the proton in $m s^{- 2}$ is

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(6i^−9j^+4k^)×9.6×1012

(6i^+9j^+4k^)×9.6×1012

(6i^−9j^−4k^)×9.6×1012

(6i^+9j^−4k^)×9.6×1012

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detailed solution

Correct option is C

q(v→×B→)=mv2r or acceleration =v2rr=qm(v→×B→) =9.6×107×[(3i^+2j^)105×(2i^+3k^)] =9.6×1012×[6i^−9j^−4k^]ms−2

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