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A charge particle is moving in a uniform magnetic field $\mathrm{(2 \hat{i}+3 \hat{j})\; \mathrm{T}.}$ If it has an acceleration of $\mathrm{(\alpha \hat{i}-4 \hat{j}) \; ^\mathrm{m} / \mathrm{s}^{2}},$ then the value of $\mathrm{\alpha}$ will be :
Option: 1
$3$

Option: 2
$6$

Option: 3
$12$

Option: 4
$2$

Answers (1)

best_answer

$\mathrm{\bar{B}=2 \hat{i}+3 \hat{\jmath} }$

$\mathrm{\bar{a}=\alpha \hat{i}-4 \hat{\jmath} }$

In magnetic field, the force vector and magnetic field vector are perpendicular to each other
i.e. $\mathrm{ \bar{F} \perp \bar{B}}$

$\mathrm{\therefore \quad \bar{a} \perp \bar{B}}$

$\mathrm{\bar{a}\rightarrow acceleration\: vector}$

$\mathrm{ \bar{a} \cdot \bar{B}= a\: B cos\: 90^{\circ}=0}$

$\mathrm{2 \alpha-12=0}$

$\mathrm{\alpha =6}$

Hence (2) is correct option.

Posted by

Pankaj Sanodiya

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