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

Step 1: The field points along \(\hat{k}\) and depends only on \(x\): \(B(x) = B_0\left(1 + \dfrac{x}{l}\right)\). Place the loop with corners at \(x\) and \(x+l\).

Step 2: Consider the two sides parallel to the X-axis. They carry equal and opposite currents over the same range of \(x\), so the field they experience is identical point by point. Their forces are equal and opposite and cancel.

Step 3: Consider the two sides parallel to the Y-axis, located at \(x\) and \(x+l\). Each has length \(l\) and lies in a uniform field along its length. The force on a side is \(F = i\,l\,B(x)\), directed along \(\hat{x}\), with opposite signs for the two sides:
\[F_{net} = i\,l\,B(x+l) - i\,l\,B(x)\]
Step 4: Substitute the field values:
\[F_{net} = i\,l\,B_0\left[\left(1 + \frac{x+l}{l}\right) - \left(1 + \frac{x}{l}\right)\right] = i\,l\,B_0\cdot\frac{l}{l} = iB_0 l\]
\[\boxed{F_{net} = iB_0 l}\]