Step 1: Understanding the Concept:
A current-carrying straight wire placed in a uniform magnetic field experiences a magnetic force.
This force is a vector quantity that depends on the current, the length vector of the wire, and the magnetic field vector.
Step 2: Key Formula or Approach:
The magnetic force $\vec{F}$ on a straight wire is given by the cross product:
\[ \vec{F} = I(\vec{L} \times \vec{B}) \]
where $I$ is the current, $\vec{L}$ is the length vector pointing in the direction of the current, and $\vec{B}$ is the magnetic field vector.
The magnitude of the force is $|\vec{F}| = \sqrt{F_x^2 + F_y^2 + F_z^2}$.
Step 3: Detailed Explanation:
The wire lies along the x-axis, so the length vector is:
\[ \vec{L} = L\hat{i} \]
The magnetic field is given as:
\[ \vec{B} = B_0(\hat{i} - \hat{j} - \hat{k}) = B_0\hat{i} - B_0\hat{j} - B_0\hat{k} \]
Now, calculate the force vector using the cross product:
\[ \vec{F} = I [ (L\hat{i}) \times (B_0\hat{i} - B_0\hat{j} - B_0\hat{k}) ] \]
Distribute the cross product:
\[ \vec{F} = I \cdot L \cdot B_0 [ \hat{i} \times (\hat{i} - \hat{j} - \hat{k}) ] \]
\[ \vec{F} = ILB_0 [ (\hat{i} \times \hat{i}) - (\hat{i} \times \hat{j}) - (\hat{i} \times \hat{k}) ] \]
Using the properties of cross products of unit vectors ($\hat{i} \times \hat{i} = 0$, $\hat{i} \times \hat{j} = \hat{k}$, $\hat{i} \times \hat{k} = -\hat{j}$):
\[ \vec{F} = ILB_0 [ 0 - (\hat{k}) - (-\hat{j}) ] \]
\[ \vec{F} = ILB_0 ( \hat{j} - \hat{k} ) \]
The force vector has components in the y and z directions. We need to find its magnitude:
\[ |\vec{F}| = \sqrt{(ILB_0)^2 + (-ILB_0)^2} \]
\[ |\vec{F}| = \sqrt{I^2 L^2 B_0^2 + I^2 L^2 B_0^2} \]
\[ |\vec{F}| = \sqrt{2 I^2 L^2 B_0^2} \]
\[ |\vec{F}| = \sqrt{2} ILB_0 \]
Step 4: Final Answer:
The magnitude of magnetic force acting on the wire is $\sqrt{2}ILB_0$.