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(C) The magnetic force on a small current element $I d\vec{l}$ is given by $d\vec{F} = I (d\vec{l} 	imes \vec{B})$.Here, the current element is $d\ve

Vedclass · Accepted Answer

$-1$

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(C) The magnetic force on a small current element $I d\vec{l}$ is given by $d\vec{F} = I (d\vec{l} 	imes \vec{B})$.Here, the current element is $d\vec{l} = dx \hat{i}$ and the magnetic field is $\vec{B} = B_0 \left( 2 - \frac{x}{a} ight) \hat{k}$.Thus, $d\vec{F} = I (dx \hat{i}) 	imes B_0 \left( 2 - \frac{x}{a} ight) \hat{k} = I B_0 \left( 2 - \frac{x}{a} ight) dx (\hat{i} 	imes \hat{k}) = I B_0 \left( 2 - \frac{x}{a} ight) dx (-\hat{j})$.To find the total force, we integrate from $x = a$ to $x = 2a$:$\vec{F} = \int_{a}^{2a} I B_0 \left( 2 - \frac{x}{a} ight) dx (-\hat{j})$$\vec{F} = -I B_0 \hat{j} \int_{a}^{2a} \left( 2 - \frac{x}{a} ight) dx$$\vec{F} = -I B_0 \hat{j} \left[ 2x - \frac{x^2}{2a} ight]_{a}^{2a}$$\vec{F} = -I B_0 \hat{j} \left[ \left( 2(2a) - \frac{(2a)^2}{2a} ight) - \left( 2(a) - \frac{a^2}{2a} ight) ight]$$\vec{F} = -I B_0 \hat{j} \left[ (4a - 2a) - (2a - 0.5a) ight]$$\vec{F} = -I B_0 \hat{j} [ 2a - 1.5a ] = -I B_0 \left( \frac{a}{2} ight) \hat{j}$.Comparing this with the given force $\vec{F} = I B_0 \left( \frac{ka}{2} ight) \hat{j}$, we get $k = -1$.

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