 Solution

ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2013/paper-2/section-a/a1/solution

y

direction on the edge

x = a

, and in the negative

x

direction on the edge

y = a

. Use

d F = I d ℓ \times B

. The edges on the coordinate axes contribute zero because the applied field vanishes there. The other two give

F_{x = a} = I α a \int_{0}^{a} y d y \hat{x}, F_{y = a} = I α a \int_{0}^{a} x d x \hat{y} .

Thus the force on the square itself, excluding the supply leads, is

F = \frac{I α a ^{3}}{2} (\hat{x} + \hat{y}) = 2.5 (\hat{x} + \hat{y}) N, ∣ F ∣ = 3.54 N .

It lies in the plane along the diagonal towards increasing

x

and

y

, at

4 5^{\circ}

to either positive axis.

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