Question

a loop of wire carrying a current of 3.13 a is in the shape of an isosceles right - triangle (right triangle with two equal sides, each 15.0 cm long). a 9.69 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 28.6 magnetic force on a current - carrying wire

Explanation:

Step1: Recall the formula for magnetic force on a current - carrying wire

The formula for the magnetic force on a current - carrying wire is $F = ILB\sin\theta$, 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 situation for the two equal - length sides of the right - isosceles triangle

The two equal sides of the right - isosceles triangle are perpendicular to the magnetic field (since the magnetic field is parallel to the hypotenuse), so $\theta = 90^{\circ}$ and $\sin\theta=1$. The length of each of the two equal sides $L = 15.0\ cm=0.15\ m$, the current $I = 3.13\ A$, and the magnetic field $B = 9.69\ T$.

Step3: Calculate the force on one side

Using the formula $F = ILB\sin\theta$, substituting the values: $F_1=ILB=(3.13\ A)\times(0.15\ m)\times(9.69\ T)$.
$F_1 = 3.13\times0.15\times9.69=4.53\ N$.

Step4: Calculate the net force on the two sides

Since the two forces on the two equal sides are perpendicular to each other (because of the right - angled triangle geometry), and have the same magnitude $F_1 = F_2$, the magnitude of the resultant force $F_{net}$ on the two sides is given by $F_{net}=\sqrt{F_1^{2}+F_2^{2}}$. Since $F_1 = F_2$, $F_{net}=\sqrt{2}F_1$.
$F_{net}=\sqrt{2}\times4.53\ N\approx6.40\ N$.

Answer:

$6.40$