A loop of wire carrying a current of 2.0 A is in the shape of a right triangle with two equal sides, each 15 cm long. A 0.7 T uniform magnetic field is in the plane of the triangle and is perpendicular to the hypotenuse. The resultant magnetic force on the two sides has a magnitude of:
A) 0
B) 0.21 N
C) 0.30 N
D) 0.41 N
E) 0.51 N

Question

Quizplus · Accepted Answer

We can use the formula for the magnetic force on a current-carrying wire:
$F = ILB\sin	heta$
where $I$ is the current, $L$ is the length of the wire, $B$ is the magnetic field, and $	heta$ is the angle between the wire and the magnetic field.

For the two equal sides of the triangle, $	heta = 90^\circ$, so $\sin	heta = 1$. The length of each side is 15 cm, so the total length of the wire is $L = 30$ cm = 0.3 m.

The force on each of the two sides is then:
$F = ILB\sin	heta = (2.0	ext{ A})(0.3	ext{ m})(0.7	ext{ T})(1) = 0.42$ N

Since the two forces are at right angles to each other, the resultant force is the vector sum of the two forces, which has magnitude:
$F_{net} = \sqrt{F^2 + F^2} = \sqrt{2F^2} = \sqrt{2}(0.42	ext{ N}) = 0.60$ N

Therefore, the closest choice is C, 0.30 N.

A Loop of Wire Carrying a Current of 2