Magnetic Force on a Current-Carrying Wire — Question 8

Question 8

A localized field acts only on the exposed part of a complete current path

A rectangular circuit carries current I. Only a length ell of its lower straight side lies in a uniform into-page field B; the remaining circuit is outside the field. Current in the exposed segment flows rightward. Ignore fringe forces.

1. Find the net force on the complete circuit.

2. For I=2.50 A, ell=.0800 m and B=.600 T, find the force and the current needed to double it without changing the field region.

Question 8 - Solution

A localized field acts only on the exposed part of a complete current path

(a) Method and calculation

$$\vec F=I\int_{\mathrm{exposed}}d\vec\ell\times(-B\hat z)=I\ell B\hat y.$$$$\int_{\mathrm{unexposed}}d\vec\ell\times\vec B=0.$$

(b) Interpretation and further calculation

$$F=(2.50)(.0800)(.600)=.120\,\mathrm N.$$$$F\prime=2F\Rightarrow I\prime=2I=5.00\,\mathrm A.$$

Physical check

A closed loop has zero net force only when the relevant field is uniform over its entire path.