Question

A part of a wire carrying \( 2.0 \, \text{A} \) current and bent at \( 90^\circ \) at two points is placed in a region of uniform magnetic field \( \vec{B} = -0.50 \, \hat{k} \, \text{T} \), as shown in the figure. Calculate the magnitude of the net force acting on the wire.

Hint:

In uniform magnetic fields:

Net force on bent wire = force on straight end-to-end vector
Use vector cross product directly

Answer 1

Step 1: Geometry of wire.
Vertical segment length = \( 20 \, \text{cm} = 0.20 \, \text{m} \)
Horizontal displacement = \( 50 \, \text{cm} = 0.50 \, \text{m} \)
The wire forms an L-shape.
Step 2: Net displacement vector.
\[ \vec{L}_{\text{net}} = 0.50 \hat{i} - 0.20 \hat{j} \]
Step 3: Magnetic field.
\[ \vec{B} = -0.50 \hat{k} \, \text{T} \]
Step 4: Net force.
\[ \vec{F} = I (\vec{L}_{\text{net}} \times \vec{B}) = 2 \left[(0.50 \hat{i} - 0.20 \hat{j}) \times (-0.50 \hat{k})\right] \]
Step 5: Compute cross products.
\(\hat{i} \times \hat{k} = -\hat{j}, \quad \hat{j} \times \hat{k} = \hat{i}\)
\((0.50 \hat{i}) \times (-0.50 \hat{k}) = 0.25 \hat{j}\)
\((-0.20 \hat{j}) \times (-0.50 \hat{k}) = 0.10 \hat{i}\)
\[ \vec{F} = 2 (0.10 \hat{i} + 0.25 \hat{j}) = 0.20 \hat{i} + 0.50 \hat{j} \]
Step 6: Magnitude of force.
\[ F = \sqrt{0.20^2 + 0.50^2} = \sqrt{0.04 + 0.25} = \sqrt{0.29} \approx 0.54 \, \text{N} \]
Final Answer:
\[ \boxed{0.54 \, \text{N}} \]