A wire carrying a current $I$ along the positive $x$-axis has length $L$. It is kept in a magnetic field $\vec

Question

A wire carrying a current $I$ along the positive $x$-axis has length $L$. It is kept in a magnetic field $\vec{B} = (2\hat{i} + 3\hat{j} - 4\hat{k})$ T. The magnitude of the magnetic force acting on the wire is:

Accepted Answer

1.  **Identify Formula:** The magnetic force $\vec{F}$ exerted on a straight wire of length $L$ carrying current $I$ in a magnetic field $\vec{B}$ is given by the cross product:
    $$\vec{F} = I(\vec{L} 	imes \vec{B})$$
2.  **Define Vectors:**
    *   The wire lies along the positive $x$-axis, so the length vector is $\vec{L} = L\hat{i}$.
    *   The magnetic field is given as $\vec{B} = 2\hat{i} + 3\hat{j} - 4\hat{k}$.
3.  **Calculate Cross Product:**
    Substitute the vectors into the formula:
    $$\vec{F} = I [ (L\hat{i}) 	imes (2\hat{i} + 3\hat{j} - 4\hat{k}) ]$$
    Distribute the cross product (recall $\hat{i} 	imes \hat{i} = 0$, $\hat{i} 	imes \hat{j} = \hat{k}$, and $\hat{i} 	imes \hat{k} = -\hat{j}$):
    $$\vec{F} = I [ L(2)(\hat{i} 	imes \hat{i}) + L(3)(\hat{i} 	imes \hat{j}) - L(4)(\hat{i} 	imes \hat{k}) ]$$
    $$\vec{F} = I [ 0 + 3L(\hat{k}) - 4L(-\hat{j}) ]$$
    $$\vec{F} = I (3L\hat{k} + 4L\hat{j}) = IL (4\hat{j} + 3\hat{k})$$
4.  **Calculate Magnitude:**
    The magnitude of the force vector $|\vec{F}|$ is:
    $$|\vec{F}| = IL \sqrt{4^2 + 3^2}$$
    $$|\vec{F}| = IL \sqrt{16 + 9} = IL \sqrt{25}$$
    $$|\vec{F}| = 5IL$$

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