Question

The magnetic field existing in a region is given by \[\vec{B} = 0.2(1 + 2x)\hat{k} \, \text{T}.\]A square loop of edge \( 50 \, \text{cm} \) carrying \( 0.5 \, \text{A} \) current is placed in the \( x \)-\( y \) plane with its edges parallel to the \( x \)- and \( y \)-axes, as shown in the figure. The magnitude of the net magnetic force experienced by the loop is ______ \( \text{mN} \).

Answer 1

Correct Answer: 50

Magnetic Force on a Current-Carrying Wire in a Magnetic Field:
 The magnetic force \( F \) on a segment of current-carrying wire in a magnetic field is given by:
\[ F = ILB \sin \theta \] where:
\( I \) is the current,
\( L \) is the length of the segment,
\( B \) is the magnetic field,
\( \theta \) is the angle between the magnetic field and the current direction.

In this case, the loop lies in the \( x-y \) plane with its edges parallel to the \( x- \) and \( y- \) axes. Since \( \vec{B} \) varies with \( x \), the magnetic force on each side of the loop depends on its position in the \( x-y \) plane.

Calculate the Force on Each Side of the Loop:
For the left side at \( x = 0 \):
\[ B_{\text{left}} = 0.2(1 + 2 \times 0) = 0.2 \, \text{T} \]
Force on the left side:
\[ F_{\text{left}} = ILB_{\text{left}} = 0.5 \times 0.5 \times 0.2 = 0.05 \, \text{N} \]

For the right side at \( x = 0.5 \, \text{m} \):
\[ B_{\text{right}} = 0.2(1 + 2 \times 0.5) = 0.2 \times 2 = 0.4 \, \text{T} \]
Force on the right side:
\[ F_{\text{right}} = ILB_{\text{right}} = 0.5 \times 0.5 \times 0.4 = 0.1 \, \text{N} \]

Net Force on the Loop:
The forces on the top and bottom sides (parallel to the \( x \)-axis) will cancel each other out due to symmetry, as the magnetic field along these sides is the same.
Therefore, the net force is due to the difference in the forces on the left and right sides:
\[ F_{\text{net}} = F_{\text{right}} - F_{\text{left}} = 0.1 - 0.05 = 0.05 \, \text{N} \]

Convert to mN:
\[ F_{\text{net}} = 0.05 \, \text{N} = 50 \, \text{mN} \]

Conclusion:
The magnitude of the net magnetic force experienced by the loop is \( 50 \, \text{mN} \).

Answer 2

Step 1: Given information.
Magnetic field: \( \vec{B} = 0.2(1 + 2x)\hat{k} \, \text{T} \)
Square loop of side \( l = 50 \, \text{cm} = 0.5 \, \text{m} \)
Current in the loop: \( I = 0.5 \, \text{A} \)
Bottom edge of the square is at \( x = 2 \, \text{m} \), and the loop lies in the x–y plane.

Step 2: Determine the magnetic field variation across the loop.
The field varies with \( x \). For sides parallel to the y-axis (vertical sides), the magnetic field is different because \( B \propto (1 + 2x) \).
For horizontal sides (parallel to x-axis), magnetic force due to current is zero since \( \vec{B} \parallel \hat{k} \) and \( \vec{I} \parallel \hat{i} \), so \( \vec{I} \times \vec{B} \) is perpendicular but equal and opposite for top and bottom sides, canceling out.

Step 3: Consider vertical sides.
For the left vertical side, position \( x_1 = 2 \, \text{m} \).
For the right vertical side, position \( x_2 = 2 + 0.5 = 2.5 \, \text{m} \).

Magnetic field values:
\[ B_1 = 0.2(1 + 2x_1) = 0.2(1 + 4) = 1.0 \, \text{T} \] \[ B_2 = 0.2(1 + 2x_2) = 0.2(1 + 5) = 1.2 \, \text{T} \]
Step 4: Magnetic forces on vertical sides.
Force on a side carrying current \( I \), length \( l \), in magnetic field \( B \) is:
\[ F = I l B \] The directions of forces on the two vertical sides are opposite, so the net force is:
\[ F_{\text{net}} = I l (B_2 - B_1) \] Substitute the values:
\[ F_{\text{net}} = 0.5 \times 0.5 \times (1.2 - 1.0) = 0.25 \times 0.2 = 0.05 \, \text{N} \] \[ F_{\text{net}} = 50 \, \text{mN} \]
Step 5: Final Answer.
\[ \boxed{F_{\text{net}} = 50 \, \text{mN}} \]