Question

The magnetic field that exists in a region is given by \(\vec{B} = B_0\left[1 + \dfrac{x}{l}\right]\hat{k}\). A square loop of edge \(l\) carrying a current \(i\) is placed with its edges parallel to the X and Y axes. The magnitude of the net magnetic force experienced by the loop is:

• A. \(2iB_0 l\)
• B. \(4iB_0 l\)
• C. \(5iB_0 l\)
• D. \(iB_0 l\)

Hint:

Sides along the field-gradient direction cancel; the net force comes from the two sides where the field differs across the loop width \(l\).

Answer 1

The Correct Option is D

Idea: Only the gradient of the field produces a net force on a current loop. Since $B$ varies linearly in $x$, only the two sides oriented so that current flows along $\hat{y}$ feel a net imbalance.

Working: For a segment carrying current $i$ along $\hat{y}$ in field $B\hat{k}$, the force per unit length is $i\,\hat{y}\times B\hat{k} = iB\,\hat{x}$. The right side sits at $x+l$ with field $B_0(1+\tfrac{x+l}{l})$ and the left side at $x$ with field $B_0(1+\tfrac{x}{l})$, carrying opposite currents. The difference in force magnitude is
\[\Delta F = i\,l\,\big[B_0(1+\tfrac{x+l}{l}) - B_0(1+\tfrac{x}{l})\big]\]
\[\Delta F = i\,l\,B_0\cdot\frac{(x+l)-x}{l} = i\,l\,B_0 = iB_0 l\]
The two sides carrying current along $\hat{x}$ experience identical opposing forces (same $x$-range) and cancel exactly.
\[\boxed{F_{net} = iB_0 l}\]