The diagram shows the arrangement you used to study forces between current carry
ID: 2058977 • Letter: T
Question
The diagram shows the arrangement you used to study forces between current carrying wires. With no current in either coil, the electronic balance shows a value of 185.12g. The current is set to 0.3A in each coil, and the balance reading changes by 0.75g. The current goes from A to B across the top segment of the top coil and from B to A across the top segment of the balance coil.
A) Determine the new balance reading and calculate the force the top coil exerts on the balance coil. Is the force toward the top of the page or toward the bottom of the page?
B) The current in the top coil is reversed and the current in each coil is increased to 0.5A. Determine the new balance reading. Use g=9.8 m/s2.
Explanation / Answer
Part A)
The bottom wire in the top coil and the top wire in the balance coil will have currents in the same direction. The force exerted between wires carrying currents in the same direction is attractive, so the lower wire will pull up on the balance coil. It can also be shown that the similar forces of attraction and repulsion will occur between other elements of the coils, but since the two wire I pointed out are closest together, it can easily be seen that the net force will be attractive and therefore reduce the amount of weight sitting on the balance.
Therefore, the reading change of .75 g, means that the new reading is
185.12 - .75 = 184.37g so the attractive force pulls up on the balance coil.
Part B)
The force exerted in part A will be equal to the amount of weight it reduced
F = (7.5 X 10-4)(9.8) = 7.35 X 10-3 N
When the current is reversed, the force will be replusive, so the scale reading will go up. We can find the force by a ratio
7.35 X 10-3/.3 A = F/.5 A
F = .01225 N
Since F = mg, the mass that involves id
.01225/9.8 = 1.25 X 10-3 kg which is 1.25 g
The new balance reading will be
185.12 + 1.25 = 186.37 g