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A wire of mass 100 g is carrying a current of 2 A towards increasing $x$ in the form of $y = x^{2}$ where $- 2 m \leq x \leq + 2 m$ . This wire is placed in a magnetic field $\vec{B} = - 0.02 \overset{\land}{K}$ tesla. The acceleration of the wire (in $m s^{- 2}$ ) is

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$- 1.6 \overset{\land}{j}$

$- 3.2 \overset{\land}{j}$

$1.6 \overset{\land}{j}$

ZERO

answer is C.

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Detailed Solution

Magnetic force on POQ is equal to magnetic force on straight wire PQ having the same current. Hence, $\vec{F} = i ({\vec{l}}_{e f f} \times \vec{B})$

$\Rightarrow \vec{F} = 2 [(4 \overset{\land}{i}) \times (- 0.02 \overset{\land}{k})] = 0.16 \overset{\land}{j}$

$\Rightarrow \vec{a} = \frac{\vec{F}}{m} = \frac{(0.16 \overset{\land}{j})}{0.1} = 1.6 \overset{\land}{j}$

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