The diagram shows the arrangement used to study forces between current-carrying
ID: 1795604 • Letter: T
Question
The diagram shows the arrangement used to study forces between current-carrying wires. With no current in either coil, the electronic balance shows a value of 0.0750 g. When a current of 0.558 A exists in each coil, the electronic balance shows a value of-0.857 g. The current in the top coil across the top segment is from A to B (a) What is the direction of the current across the top segment in - the balance coil (A to B or B to A)? (b) Calculate the magnitude of the force exerted on the balance coil by the top coil. Is the force repulsive or attractive? (c) What is the reading on the balance if the current in top coil is reversed? (d) Calculate the product of currents 1112 that would be required to produce a force of one newton between the top coil and the balance coil. (e) What is the direction of the magnetic field at point D? Please use g = 9.80 m/s Top Coil Balance Coll Balance PanExplanation / Answer
(a) if currents are in the same direction in two wires then attraction takes place between the wires.
Since it is given that current in top segment in top coil is from A to B,in the bottom segment it must be from B to A( to the left). Hence current in top segment in balance coil must also flow to the left ( From A to B)
(b) Since balance shows -0.857 g with current on and 0.0750 with no current,total attractive force on balance coil
= 0.857 + 0.075 = 0.932 g = 9.13 N
(c) if current in top coil is reversed there will be repusion between the coils 9.13 N
Since zero current reading in balance is 0.0750
total reading = 0.932 g+ 0.0750 g = 1.007 g
(d) since force is directly proportinal to I1 X I2
F1/F2 = 9.13N / 1N = (0.558A)^2/ (I1 X I2)
I1 X I2 = (1/9.13) X (0.558A)^2 = 0.034 ampere^2
(e) B is from inside to outside of the screen ( apply right hand thumb rule -thumb points along the current direction- the curled other fingers show direction of B field)