Two straight rods 60 cm long and 2.0 mm apart in a current balance carry currents of 18 A each in opposite directions.What mass must be placed on the upper rod to balance the magnetic force of repulsion?
A)0.50 g
B)0.99 g
C)9.7 g
D)4.3 g
E)1.6 g

Question

Quizplus · Accepted Answer

The magnetic force between two parallel currents can be calculated using the formula $F = \frac{\mu_0 I_1 I_2 L}{2\pi d}$, where $F$ is the force, $\mu_0$ is the permeability of free space ($4\pi \times 10^{-7} \, \text{Tm/A}$), $I_1$ and $I_2$ are the currents, $L$ is the length of the rods, and $d$ is the distance between them. Given that $I_1 = I_2 = 18 \, \text{A}$, $L = 0.6 \, \text{m}$, and $d = 2.0 \times 10^{-3} \, \text{m}$, we can calculate the force as follows:$$F = \frac{(4\pi \times 10^{-7}) \times (18)^2 \times 0.6}{2\pi \times 2.0 \times 10^{-3}} = \frac{4 \times 18^2 \times 0.6 \times 10^{-7}}{4 \times 10^{-3}} = \frac{18^2 \times 0.6 \times 10^{-7}}{10^{-3}} = \frac{324 \times 0.6 \times 10^{-7}}{10^{-3}} = 1.944 \times 10^{-4} \, \text{N}$$To balance this magnetic force, the gravitational force ($F_g$) acting on the mass ($m$) placed on the upper rod must be equal to the magnetic force. The gravitational force is given by $F_g = mg$, where $g$ is the acceleration due to gravity ($9.8 \, \text{m/s}^2$). Setting $F_g = F$ and solving for $m$ gives:$$1.944 \times 10^{-4} = m \times 9.8$$$$m = \frac{1.944 \times 10^{-4}}{9.8} = 1.984 \times 10^{-5} \, \text{kg}$$$$m = 19.84 \times 10^{-3} \, \text{g}$$$$m \approx 0.99 \, \text{g}$$Therefore, a mass of approximately 0.99 g must be placed on the upper rod to balance the magnetic force of repulsion.

Two Straight Rods 60 Cm Long and 2