The magnetic field existing in a region is given by B⃗=0.2(1+2 x) k̂. A square loop of edge 50 cm carrying 0.5 A current is placed in x-y plane with its edges parallel to the x-y axes, as shown in figure. The magnitude of the net magnetic force experienced by the loop is _________ mN.

Solution

Physics sir · 4 steps

The field changes only with x, so top and bottom forces cancel; the two vertical sides decide everything.

1. Read the loop. The loop runs from x = 2 m to x = 2.5 m and from y = 2 m to y = 2.5 m, with current going anticlockwise. Each edge feels F⃗ = IL⃗ × B⃗.
2. Horizontal edges cancel. The bottom and top edges cover the same range of x, so they see the same field. Their currents are opposite, so their forces are equal and opposite and add to zero.

   Edge	Current direction	Field B	Force
   Right, x = 2.5 m	+ĵ	0.2(1 + 5) = 1.2 T	0.5 × 0.5 × 1.2 = 0.3 N along +î
   Left, x = 2 m	-ĵ	0.2(1 + 4) = 1.0 T	0.5 × 0.5 × 1.0 = 0.25 N along -î

3. Directions. On the right edge ĵ × k̂ = î, and on the left edge (-ĵ) × k̂ = -î. The two forces point opposite ways, so they subtract.

   Fₙₑₜ = IL (B_(right) - B_(left)) = 0.5 × 0.5 × (1.2 - 1.0) = 0.05 N

4. So. The net force is 0.05 N along +î, which is 50 mN.

The common slip: 550. Adding the two side forces instead of subtracting them gives 0.3 + 0.25 = 0.55 N, that is 550 mN. The forces point in opposite directions.

The idea to keep. In a uniform field a closed loop feels zero net force. If the field varies, pair up the opposite edges: only the edges that sit at different field strengths leave a leftover force.

Answer from the NTA final answer key. Solution by PrepWiser.