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 ______ \( \text{mN} \).
Magnetic Force on a Current-Carrying Wire: The magnetic force \( F \) on a segment of wire carrying current in a magnetic field is determined by: \[ F = ILB \sin \theta \] where \( I \) denotes the current, \( L \) the segment length, \( B \) the magnetic field strength, and \( \theta \) the angle between the magnetic field and the current's direction.
In this scenario, the loop is situated in the \( x-y \) plane with its sides aligned with the \( x- \) and \( y- \) axes. Given that \( \vec{B} \) is dependent on \( x \), the magnetic force acting on each side of the loop is influenced by its specific location within the \( x-y \) plane.
Force Calculation per Loop Side: For the left edge at \( x = 0 \): \[ B_{\text{left}} = 0.2(1 + 2 \times 0) = 0.2 \, \text{T} \] The force on the left edge is: \[ F_{\text{left}} = ILB_{\text{left}} = 0.5 \times 0.5 \times 0.2 = 0.05 \, \text{N} \]
For the right edge at \( x = 0.5 \, \text{m} \): \[ B_{\text{right}} = 0.2(1 + 2 \times 0.5) = 0.2 \times 2 = 0.4 \, \text{T} \] The force on the right edge is: \[ F_{\text{right}} = ILB_{\text{right}} = 0.5 \times 0.5 \times 0.4 = 0.1 \, \text{N} \]
Net Force on the Loop: The forces acting on the top and bottom sides (oriented along the \( x \)-axis) will counteract and cancel each other due to symmetry, as the magnetic field strength is consistent along these sides. Consequently, the net force is the difference between the forces on the right and left edges: \[ F_{\text{net}} = F_{\text{right}} - F_{\text{left}} = 0.1 - 0.05 = 0.05 \, \text{N} \]
