1. **Problem Statement:** Calculate the magnetic force on each side of a right-angled triangular loop carrying current $i=4.00$ A with sides $0.5$ cm, $1.20$ cm, and $1.30$ cm placed in a uniform magnetic field $B=0.75$ mT. The magnetic field is parallel to the side of length $1.30$ cm. 2. **Formula:** The magnetic force on a current-carrying wire in a magnetic field is given by $$F = iLB\sin(\theta)$$ where $L$ is the length of the wire, $B$ is the magnetic field strength, $i$ is the current, and $\theta$ is the angle between the wire and the magnetic field. 3. **Important Rules:** - If the wire is parallel to the magnetic field, $\theta=0^\circ$ and $\sin(0)=0$, so force is zero. - If the wire is perpendicular, $\theta=90^\circ$ and $\sin(90^\circ)=1$, so force is maximum. 4. **Calculations:** - (a) Side length $1.30$ cm: Since $B$ is parallel to this side, $\theta=0^\circ$. $$F_a = 4.00 \times 0.0130 \times 0.00075 \times \sin 0^\circ = 0$$ - (b) Side length $0.5$ cm: This side is perpendicular to the $1.30$ cm side in a right triangle, so $\theta=90^\circ$. $$F_b = 4.00 \times 0.0050 \times 0.00075 \times \sin 90^\circ = 4.00 \times 0.0050 \times 0.00075 = 1.5 \times 10^{-5}$$ N - (c) Side length $1.20$ cm: This side is the other leg of the right triangle, so $\theta=90^\circ$. $$F_c = 4.00 \times 0.0120 \times 0.00075 \times \sin 90^\circ = 4.00 \times 0.0120 \times 0.00075 = 3.6 \times 10^{-5}$$ N - (d) Net force on the loop: Forces on sides (b) and (c) are perpendicular and can be combined vectorially. $$F_{net} = \sqrt{(F_b)^2 + (F_c)^2} = \sqrt{(1.5 \times 10^{-5})^2 + (3.6 \times 10^{-5})^2} = 3.9 \times 10^{-5}$$ N 5. **Explanation:** The side parallel to the magnetic field experiences no force because the magnetic force depends on the sine of the angle between current and field. The other two sides, being perpendicular, experience forces proportional to their lengths. The net force is the vector sum of these two forces. **Final answers:** - (a) $F_a = 0$ N - (b) $F_b = 1.5 \times 10^{-5}$ N - (c) $F_c = 3.6 \times 10^{-5}$ N - (d) $F_{net} = 3.9 \times 10^{-5}$ N