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(C) The magnetic force on a current-carrying wire in a uniform magnetic field is given by $\vec{F} = I (\vec{L}_{eff} 	imes \vec{B})$, where $\vec{L}

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$I \lambda B$

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(C) The magnetic force on a current-carrying wire in a uniform magnetic field is given by $\vec{F} = I (\vec{L}_{eff} 	imes \vec{B})$, where $\vec{L}_{eff}$ is the effective displacement vector from the starting point to the end point of the wire.For the given curve $y = A \sin \left(\frac{2 \pi}{\lambda} xight)$, the starting point is at $x=0$, which gives $y=0$. The end point is at $x=\lambda$, which gives $y = A \sin(2\pi) = 0$.Thus, the effective displacement vector $\vec{L}_{eff}$ is the vector from $(0, 0)$ to $(\lambda, 0)$, which is $\vec{L}_{eff} = \lambda \hat{i}$.The magnetic field is in the $z$-direction, so $\vec{B} = B \hat{k}$.The magnetic force is $\vec{F} = I (\lambda \hat{i} 	imes B \hat{k}) = I \lambda B (\hat{i} 	imes \hat{k}) = -I \lambda B \hat{j}$.The magnitude of the magnetic force is $|\vec{F}| = I \lambda B$.

$A$ wire carrying a steady current $I$ is kept in the $x$-$y$ plane along the curve $y=A \sin \left(\frac{2 \pi}{\lambda} x\right)$. $A$ magnetic field $B$ exists in the $z$-direction. The magnitude of the magnetic force on the portion of the wire between $x=0$ and $x=\lambda$ is

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