Two parallel wires carry 1-A currents in unknown directions. The distance between the wires is 10-cm. What is the magnitude of the magnetic field B in Teslas at a point P located 6-cm away from the axis of one of the wires and 8-cm away from the axis of the other wire?
Details and assumptions
\frac{\mu_{0}}{4\pi}= 10^{-7} H/m
To find the magnitude of the magnetic field at point P, we can use the Biot-Savart law. The Biot-Savart law states that the magnetic field created by a current-carrying wire at a given point is directly proportional to the current, the length of the wire, and the sine of the angle between the wire and the line connecting the wire to the point.
The formula for the magnetic field due to a current in a straight wire is:
B = \frac{\mu_{0} \cdot I}{2\pi \cdot r}
Where:
B is the magnetic field in Teslas
\mu_{0} is the permeability of free space and is equal to 4\pi \times 10^{-7} H/m
I is the current in amperes
r is the distance from the wire in meters
In this case, we have two parallel wires carrying 1-A currents in unknown directions. We need to find the magnetic field at point P, which is located 6-cm away from the axis of one wire and 8-cm away from the axis of the other wire.
First, let's calculate the magnetic field due to the first wire at point P:
B1 = \frac{\mu_{0} \cdot I1}{2\pi \cdot r1}
Where:
B1 is the magnetic field due to the first wire at point P
I1 is the current in the first wire (1 A)
r1 is the distance from the first wire to point P (6 cm = 0.06 m)
Now, let's calculate the magnetic field due to the second wire at point P:
B2 = \frac{\mu_{0} \cdot I2}{2\pi \cdot r2}
Where:
B2 is the magnetic field due to the second wire at point P
I2 is the current in the second wire (1 A)
r2 is the distance from the second wire to point P (8 cm = 0.08 m)
Finally, to get the total magnetic field at point P, we need to find the vector sum of B1 and B2:
B = \sqrt{B1^2 + B2^2}
Using the given values and the formulas above, we can calculate the magnitude of the magnetic field at point P.