Question

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 0.70 t uniform magnetic field is parallel to the hypotenuse. the total magnetic force on the two equal sides has a magnitude of
o 0.41 n.
o 0.51 n.
o 0.21 n.
o 0.0 n.
o 0.30 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 situation for the two equal - length sides

For each of the two equal sides of length $L = 15\ cm=0.15\ m$, $I = 2.0\ A$, $B = 0.70\ T$. Since the magnetic field is parallel to the hypotenuse of the right - triangle loop, the angle $\theta$ between the magnetic field and each of the two equal sides is $90^{\circ}$, and $\sin\theta=\sin90^{\circ}=1$.

Step3: Calculate the force on one side

Using the formula $F = ILB\sin\theta$, substituting the values: $F_1=ILB=(2.0\ A)\times(0.15\ m)\times(0.70\ T)=0.21\ N$.

Step4: Calculate the total force on the two equal sides

Since the two equal sides have the same force acting on them and the forces are perpendicular to each other (because of the right - triangle shape of the loop), the total force on the two equal sides is $F_{total}=\sqrt{F_1^{2}+F_1^{2}}=\sqrt{2}F_1$. Substituting $F_1 = 0.21\ N$, we get $F_{total}=\sqrt{2}\times0.21\ N\approx0.30\ N$.

Answer:

0.30 N