Question

a loop of wire carrying a current of 7.89 a is in the shape of an isosceles right triangle (right triangle with two equal sides, each 15.0 cm long). a 1.35 t uniform magnetic field is parallel to the hypotenuse of the right triangle. what is the magnitude of the resulting magnetic force, in newtons, on the two sides? n

Explanation:

Step1: Recall magnetic - force formula

The magnetic force on a current - carrying wire is given by $F = ILB\sin\theta$, where $I$ is the current, $L$ is the length of the wire, $B$ is the magnetic field, and $\theta$ is the angle between the wire and the magnetic field.

Step2: Analyze the angle for the two sides

Since the magnetic field is parallel to the hypotenuse of the right - isosceles triangle, for the two equal sides of length $L = 15.0\ cm=0.15\ m$, the angle $\theta = 45^{\circ}$, and $\sin\theta=\sin45^{\circ}=\frac{\sqrt{2}}{2}$. The current $I = 7.89\ A$ and the magnetic field $B = 1.35\ T$.

Step3: Calculate the force on one side

Using the formula $F = ILB\sin\theta$, we substitute the values: $F_1=ILB\sin\theta=(7.89\ A)\times(0.15\ m)\times(1.35\ T)\times\frac{\sqrt{2}}{2}$.
$F_1=(7.89)\times(0.15)\times(1.35)\times\frac{\sqrt{2}}{2}\ N\approx1.11\ N$.

Step4: Calculate the net force on the two sides

Since the two forces on the two equal sides are perpendicular to each other (because it's a right - isosceles triangle), the magnitude of the net force $F_{net}=\sqrt{F_1^{2}+F_1^{2}}=\sqrt{2}F_1$.
$F_{net}=\sqrt{2}\times1.11\ N\approx1.57\ N$.

Answer:

$1.57$