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Step-by-Step Solution
Step 1: Write down the formula for magnetic force
The magnetic force on a current-carrying wire is given by the vector cross product:
$$
\vec{F} = I \bigl(\vec{\ell} \times \vec{B}\bigr),
$$
where $I$ is the current, $\vec{\ell}$ is the length vector of the wire (in the direction of current), and $\vec{B}$ is the magnetic field.
Step 2: Identify the vectors
We have:
– Current $I$ along the positive $x$-axis, so $\vec{\ell} = L \,\hat{i}$.
– Magnetic field $\vec{B} = 2 \,\hat{i} + 3 \,\hat{j} - 4 \,\hat{k}$.
Step 3: Calculate the cross product $\vec{\ell} \times \vec{B}$
$$
\vec{\ell} \times \vec{B} \;=\;
\bigl(L \,\hat{i}\bigr) \times \bigl(2 \,\hat{i} + 3 \,\hat{j} - 4 \,\hat{k}\bigr).
$$
Recall the cross product properties:
– $\hat{i} \times \hat{i} = 0$,
– $\hat{i} \times \hat{j} = \hat{k}$,
– $\hat{i} \times \hat{k} = -\,\hat{j}.$
Therefore:
$$
\begin{aligned}
L \,\hat{i} \times (2 \,\hat{i}) & = 0, \\
L \,\hat{i} \times (3 \,\hat{j}) & = 3L \,\hat{i} \times \hat{j} = 3L \,\hat{k}, \\
L \,\hat{i} \times \bigl(-4 \,\hat{k}\bigr)
& = -4L \Bigl(\hat{i} \times \hat{k}\Bigr)
= -4L \, \bigl(-\hat{j}\bigr)
= 4L \,\hat{j}.
\end{aligned}
$$
Hence,
$$
\vec{\ell} \times \vec{B} = 4L \,\hat{j} \;+\; 3L \,\hat{k}.
$$
Step 4: Multiply by the current $I$ and find the force vector
$$
\vec{F} = I(\vec{\ell} \times \vec{B})
= I \Bigl(4L \,\hat{j} + 3L \,\hat{k}\Bigr).
$$
Step 5: Calculate the magnitude of the force
The magnitude of the force $\bigl|\vec{F}\bigr|$ is given by the magnitude of the vector $4L \,\hat{j} + 3L \,\hat{k}$ multiplied by $I$:
$$
\bigl|\vec{F}\bigr|
= I \sqrt{\bigl(4L\bigr)^2 + \bigl(3L\bigr)^2}
= I \sqrt{16L^2 + 9L^2}
= I \sqrt{25L^2}
= 5\,I\,L.
$$
Step 6: State the final answer
The magnitude of the magnetic force acting on the wire is $5\,IL$.
