Two very long, straight, parallel wires carry currents that are directed perpendicular. Wire 1 carries a current I1 into the page (in the -z direction) and passes through the x axis at x = +a. Wire 2 passes through the x axis at x = -2a and carries an unknown current I2. The total magnetic field at the origin due to the current-carrying wires has the magnitude 2µ0I1 / ( 2πa ). The current I2 can have either of two possible values.
Find the value of I2 with the smaller magnitude, stating it in terms of I1 and giving its direction.
I have to write formula
thank you
Not exactly write a formula. You have to analyze the problem, then derive the formula. We don't write formulas in Physics, the Creator did that.
Here first find the distance between wires (so what is the distance between x=-2a and x=a) The force at the origin? Well it is the sum of the two forces from each wire.
Force from wire1 is kI/(2pi*distance)
http://dev.physicslab.org/Document.aspx?doctype=3&filename=Magnetism_CurrentCarryingWires.xml
The other force is similar, but the distances are different.
Now signs. Use your right hand rule, and determine the force direction for each wire. You don't know the direction for current I2, but if it is into the page, then the force subtracts from the first force.
Well, add the forces, with the +- force for the second force.
To find the value of I2 with the smaller magnitude, we can use the principle of superposition in magnetism. This principle states that the total magnetic field at a point due to multiple current-carrying wires can be found by summing up the individual magnetic fields produced by each wire.
Given that the total magnetic field at the origin (0, 0) is given by:
B_total = 2µ0I1 / (2πa) ----(1)
Where:
B_total = Total magnetic field at the origin
µ0 = Permeability of free space (constant)
I1 = Current in wire 1
a = Distance of wire 1 from the origin
We can calculate the individual magnetic fields produced by each wire and then equate it to equation (1).
For wire 1:
The magnetic field produced by wire 1 can be found using Ampere's law, given by:
B_1 = (µ0 * I1) / (2π * d1) ----(2)
Where:
B_1 = Magnetic field due to wire 1
d1 = Distance of the origin from wire 1 (d1 = a)
For wire 2:
The magnetic field produced by wire 2 can be found using Ampere's law, given by:
B_2 = (µ0 * I2) / (2π * d2) ----(3)
Where:
B_2 = Magnetic field due to wire 2
d2 = Distance of the origin from wire 2 (d2 = 2a)
Since the wires are parallel and carry currents perpendicular to each other, the net magnetic field at the origin will be the vector sum of the magnetic fields due to each wire. Since their directions are opposite, we need to subtract the magnitudes when calculating the net magnetic field.
Therefore, the net magnetic field at the origin is given by:
B_total = |B_1 - B_2|
Substituting equations (2) and (3) into the above expression, we get:
2µ0I1 / (2πa) = |(µ0 * I1) / (2πa) - (µ0 * I2) / (2π * 2a)|
Simplifying this equation, we find:
|I1 - I2 / 2| = 1
Now, we can solve for I2:
I1 - I2 / 2 = 1 or -(I1 - I2 / 2) = 1
Solving each equation, we get:
I2 = I1 - 2 or I2 = 3I1 - 2
Therefore, the two possible values of I2 are I1 - 2 and 3I1 - 2. Since we want the value of I2 with the smaller magnitude, we choose I2 = I1 - 2.
To solve this problem, we can use the formula for the magnetic field at a point on the axis of a current-carrying wire, given by:
B = (µ0 * I) / (2π * r)
Where B is the magnetic field, µ0 is the permeability of free space, I is the current, and r is the distance from the wire.
Since the direction of the current in wire 1 is into the page (-z direction), the magnetic field it produces will have a direction in the +y direction.
Now let's consider the magnetic field at the origin due to wire 1. The distance from wire 1 to the origin is a. Using the formula above, we can write:
B1 = (µ0 * I1) / (2π * a) (in the +y direction)
Next, let's consider the magnetic field at the origin due to wire 2. The distance from wire 2 to the origin is 2a. Using the same formula, we can write:
B2 = (µ0 * I2) / (2π * 2a) (in the -y direction)
The total magnetic field at the origin due to both wires is given as 2µ0I1 / (2πa).
Since the magnetic fields from both wires add up, we can write:
B_total = B1 + B2 = 2µ0I1 / (2πa)
Substituting the expressions for B1 and B2, we get:
(µ0 * I1) / (2π * a) + (µ0 * I2) / (2π * 2a) = 2µ0I1 / (2πa)
Simplifying the equation:
I2 / 2 = I1
Dividing both sides by 2, we find:
I2 = 2I1
So, the value of I2 with the smaller magnitude is 2I1. Its direction is in the opposite direction of I1.