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A straight wire is placed in a magnetic field that varies with distance $x$ from origin as $\vec{B}=B_0\left(2-\frac{x}{a}\right) \hat{k}$. Ends of wire are at $(a, 0)$ and $(2 a, 0)$ and it carries a current $\mathrm{I}$. If force on wire is $\vec{F}=\mathrm{IB}_0\left(\frac{k a}{2}\right) \hat{\mathrm{j}}$, then value of $k$ is
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The correct answer is:
$-1$
$\overrightarrow{\mathrm{dF}}=\mathrm{Idx} \times \overrightarrow{\mathrm{B}}=\operatorname{Idx} \hat{\mathrm{i}} \times \mathrm{B} \hat{\mathrm{k}}=\operatorname{IdxB}(-\hat{\mathrm{j}})$
$\int \mathrm{dF}=\int_{\mathrm{a}}^{2 a} \mathrm{Id} x \mathrm{~B}_0\left(2-\frac{\mathrm{x}}{\mathrm{a}}\right)(-\hat{\mathrm{j}})$
$$
\begin{aligned}
&\mathrm{F}=\mathrm{IB}_0 \int_{\mathrm{a}}^{2 \mathrm{a}}\left(2-\frac{\mathrm{x}}{\mathrm{a}}\right) \mathrm{dx}(-\hat{\mathrm{j}}) \\
&=\operatorname{IB}_0\left(2 \mathrm{x}-\frac{\mathrm{x}^2}{2 \mathrm{a}}\right)_{\mathrm{a}}^{2 \mathrm{a}}(-\hat{\mathrm{j}}) \\
&=\mathrm{IB}_0\left[\left(2 \times 2 \mathrm{a}-\frac{(2 \mathrm{a})^2}{2 \mathrm{a}}\right)-\left(2 \mathrm{a}-\frac{\mathrm{a}^2}{2 \mathrm{a}}\right)\right](-\hat{\mathrm{j}}) \\
&=\mathrm{IB}_0\left[(4 \mathrm{a}-2 \mathrm{a})-\left(\frac{3 \mathrm{a}}{2}\right)\right](-\hat{\mathrm{j}}) \\
&=\mathrm{IB}_0\left[2 \mathrm{a}-\frac{3 \mathrm{a}}{2}\right](-\hat{\mathrm{j}}) \\
&=\operatorname{IB}_0\left[\frac{\mathrm{a}}{2}\right](-\hat{\mathrm{j}})
\end{aligned}
$$
Comparing $\mathrm{F}$ with $\mathrm{IB}_0\left[\frac{\mathrm{ka}}{2}\right] \hat{\mathrm{j}}$, we get $\mathrm{k}=-1$
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