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A wire of mass 100 g carrying a current of 2 A towards increasing x is in the form of y =
${x^2}( - 2m leqslant x leqslant + 2m)$
.This wire is placed in a magnetic field
$B= - 0.02{text{ }}hat k$
tesla. The acceleration of the wire (in m / s² ) is

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$- 1.6hat j$

$- 3.2hat j$

$1.6hat j$

$2.4hat j$

answer is C.

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Detailed Solution

$\vec{F} = i \vec{l} \times \vec{B}$

Acceleration $= \frac{F}{m} = \frac{i l B}{m} = \frac{2 \times 4 \times 0.02}{0.1} = 1.6 m / s^{2}$

$\vec{a} = 1.6 \hat{j}$

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