AP Physics C E and M

Question

Two parallel wires in the figure are separated by a distance a = 0.100 m and carry currents I₁ =3.00 A and I₂ =1.00 A in opposite directions.

The point A is midway between the wires.

1. What is the direction of the magnetic field B₁ at the point A created by just the current I₁ ?

In

Out

Up

Down

Left

Right

2. What is the direction of the magnetic field B₂ at the point A created by just the current I₂?

In

Out

Up

Down

Left

Right

3. What is the magnitude of the magnetic field B in [µT] at the point A due to both currents?

4 . What is the direction of the force created by the current I₁ on the wire carrying current I₂?

In

Out

Up

Down

Left

Right

5 . What is the magnitude of the force in [µN] that current I₁ exerts on a 0.100 m length of the wire carrying current I₂?

Answer 1

1.) In

The direction of the magnetic field due to a straight wire is found with the right-hand rule. Line up your right hand in the direction of the flow of current with your thumb sticking out. Then curl your fingers so that they point at the spot you are examining (i.e. point A) along the shortest possible route. Your thumb is pointing in the direction of the field, in this case, into the page.

2.) In

Another simple application of the right-hand rule. Alternatively, you could recognize that point A is on the opposite side of wire 2 as it was to wire 1, which should flip the direction of the field, but wire 2 also has the current flowing in the opposite direction, which flips the direction again. The result is that the direction remains the same as for wire 1.

3.) B=(u0/(pi a))(I1+I2)=16e-6 T=1.6e-5T

Recall the magnetic field due to a wire is B=u0 I/(2 pi r), where u0 is the permeability of free space constant, I is the current, and r is the perpendicular distance from the wire to the point of interest. Magnetic fields super impose, so we can simply find the magnetic field of each wire separately, and then add those together to get the total magnetic field. This is made even easier since both point in the same direction.

The value for r is the same for both wires, since point A is halfway between them. Since the distance between the wires is a, the distance from each wire to point A is 1/2 a. Therefore

B1=u0/(pi a) I1

B2=u0/(pi a) I2

Adding these together and factoring out the factors common to both, we get

B=B1+B2=u0/(pi a) I1+u0/(pi a) I1=(u0/(pi a)) (I1+I2)

Substitute the values for a, I1, I2, and u0 to get the numerical value.

4.) Up

We again rely on the right hand rule, but this time as applied to the force on a wire. Line your right hand up with the direction of current in the wire with your thumb outstretched, and then curl your fingers in the direction of the magnetic field acting on the wire. Your thumb points in the direction of the force.

The trick here is to recognize what the direction of the field is, but that just amounts to doing the right hand rule for magnetic fields again (or recognizing that the direction will be the same as point A, since wire 1 is in the same direction relative to wire 2 as point A) to find that the direction is IN to the page. So, we find the magnetic force points up.

5.) u0 L I1 I2/(2pi a)=0.6uN

Since the direction of the current in wire 1 is perpendicular to the magnetic field generated by wire 2, we can use the equation F=ILB, where I is the current in the wire (specifically wire 1), L is the length of wire, and B is the magnetic field acting on the wire (in this case, supplied by wire 2). L is given to be 0.1m, and the current I=I1=3A is also given. We just need to find the magnetic field.

This is the same process as for part 3. The only differences are that 1.) We only need the magnetic field from wire 2, since wire 1 can't act on itself, and 2.) The distance is different. The distance now is the full distance between the two wires, or a. Therefore

B=u0/(2pi a)I2

And so we have

F=ILB=I1 L u0/(2pi a)I2=u0 L I1 I2/(2pi a)=0.6uN