Two long, parallel conductors, separated by 14.0 cm, carry currents in the same direction. The first wire carries current I1 = 6.00 A, and the second carries I2 = 8.00 A Assume the conductors lie in the plane of the page.
(a) What is the magnitude of the magnetic field created by I1 at the location of I2?
T
(b) What is the force per unit length exerted on I2 by I1?
N/m
(c) What is the magnitude of the magnetic field created by I2 at the location of I1?
T
(d) What is the force per length exerted by I2 on I1?
N/m
direction
I₁=6A, I₂=8 A, b=0.14 m.
(a) B₁=μ₀I₁/2πb=4π•10⁻⁷•6/2π•0.14=8.57•10⁻⁶ T
(b)
F₂ =I₂B₁Lsinα
α=90° =>sinα=1
F₂/L =I₂B₁= =μ₀I₁I₂/2πb=8•8.57•10⁻⁶=6.86•10⁻⁵N/m
(c)
B₂=μ₀I₂/2πb=4π•10⁻⁷•8/2π•0.14=1.14•10⁻⁵ T
(d)
F₁ =I₁B₂Lsinα
α=90° =>sinα=1
F₁/L =I₁B₂=μ₀I₁I₂/2πb=
=6•1.14•10⁻ ⁵=6.86•10⁻⁵N/m
I₁↑ F₁→ ←F₂↑I₂
To find the answers, we can use the Biot-Savart law and the magnetic force formula.
(a) The magnitude of the magnetic field created by I1 at the location of I2 can be found using the Biot-Savart law:
B1 = (μ0 * I1) / (2π * r1)
where μ0 is the permeability of free space (4π × 10^-7 T·m/A), I1 is the current in the first wire (6.00 A), and r1 is the distance between the wires (14.0 cm = 0.14 m).
Plugging in the values:
B1 = (4π × 10^-7 T·m/A) * (6.00 A) / (2π * 0.14 m)
B1 = 8.57 × 10^-6 T
Therefore, the magnitude of the magnetic field created by I1 at the location of I2 is 8.57 × 10^-6 T.
(b) The force per unit length exerted on I2 by I1 can be calculated using the magnetic force formula:
F12 = (μ0 * I1 * I2 * L) / (2π * d)
where μ0 is the permeability of free space (4π × 10^-7 T·m/A), I1 is the current in the first wire (6.00 A), I2 is the current in the second wire (8.00 A), L is the length of the second wire, and d is the distance between the wires.
Since the conductors are long, we can assume L is large. Let's assume L = 1 m for simplicity.
Plugging in the values:
F12 = (4π × 10^-7 T·m/A) * (6.00 A) * (8.00 A) * (1 m) / (2π * 0.14 m)
F12 = 6.86 N/m
Therefore, the force per unit length exerted on I2 by I1 is 6.86 N/m.
(c) The magnitude of the magnetic field created by I2 at the location of I1 can be found using the Biot-Savart law similarly to part (a):
B2 = (μ0 * I2) / (2π * r2)
where μ0 is the permeability of free space (4π × 10^-7 T·m/A), I2 is the current in the second wire (8.00 A), and r2 is the distance between the wires (14.0 cm = 0.14 m).
Plugging in the values:
B2 = (4π × 10^-7 T·m/A) * (8.00 A) / (2π * 0.14 m)
B2 = 11.4 × 10^-6 T
Therefore, the magnitude of the magnetic field created by I2 at the location of I1 is 11.4 × 10^-6 T.
(d) The force per unit length exerted on I1 by I2 can be calculated using the magnetic force formula similarly to part (b):
F21 = (μ0 * I1 * I2 * L) / (2π * d)
Using the same assumptions as before, let's assume L = 1 m.
Plugging in the values:
F21 = (4π × 10^-7 T·m/A) * (6.00 A) * (8.00 A) * (1 m) / (2π * 0.14 m)
F21 = 8.57 N/m
Therefore, the force per unit length exerted on I1 by I2 is 8.57 N/m.
The directions of these forces can be determined using the right-hand rule.
To find the answers to these questions, we can use the Biot-Savart Law and the Lorentz Force Law, which describe the magnetic fields created by current-carrying wires and the forces between them, respectively.
(a) To find the magnitude of the magnetic field created by I1 at the location of I2, we can use the Biot-Savart Law. The formula for the magnetic field at a point due to a current-carrying wire is given by:
B = (μ₀/4π) * (I * sinθ) / r²
Where:
B is the magnetic field
μ₀ is the vacuum permeability (4π × 10^(-7) T·m/A)
I is the current
θ is the angle between the current direction and the line connecting the point and the wire
r is the distance between the point and the wire
In this case, the currents are in the same direction, so the angle θ is 0. We can assume the point where I2 is located is at a distance r from I1.
Plugging in the values:
I = 6.00 A
r = 14.0 cm = 0.14 m
B = (4π × 10^(-7) T·m/A) * (6.00 A * sin(0)) / (0.14 m)²
B ≈ 3.04 × 10^(-5) T (rounded to two decimal places)
Therefore, the magnitude of the magnetic field created by I1 at the location of I2 is approximately 3.04 × 10^(-5) T.
(b) To find the force per unit length exerted on I2 by I1, we can use the Lorentz Force Law. The formula for the force per unit length between two parallel conductors carrying currents is given by:
F/L = (μ₀/2π) * (I1 * I2 * d) / r
Where:
F/L is the force per unit length
μ₀ is the vacuum permeability (4π × 10^(-7) T·m/A)
I1 and I2 are the currents on the two conductors
d is the separation between the conductors
r is the distance between the point on one conductor and the other conductor
Plugging in the values:
I1 = 6.00 A
I2 = 8.00 A
d = 14.0 cm = 0.14 m
r = 0 (since we're looking at the force exerted on I2 by I1)
F/L = (4π × 10^(-7) T·m/A) * (6.00 A * 8.00 A * 0.14 m) / 0
F/L = 0 N/m
Therefore, the force per unit length exerted on I2 by I1 is 0 N/m. Due to the same-direction currents, the wires do not exert any net force on each other along their length.
(c) To find the magnitude of the magnetic field created by I2 at the location of I1, we can again use the Biot-Savart Law. Using similar calculations as in part (a), we can find the magnetic field due to I2 at the position of I1.
Plugging in the values:
I = 8.00 A
r = 14.0 cm = 0.14 m
B = (4π × 10^(-7) T·m/A) * (8.00 A * sin(0)) / (0.14 m)²
B ≈ 4.05 × 10^(-5) T (rounded to two decimal places)
Therefore, the magnitude of the magnetic field created by I2 at the location of I1 is approximately 4.05 × 10^(-5) T.
(d) Similar to part (b), the force per length exerted by I2 on I1 can be calculated using the Lorentz Force Law.
Plugging in the values:
I1 = 6.00 A
I2 = 8.00 A
d = 14.0 cm = 0.14 m
r = 0 (since we're looking at the force exerted on I1 by I2)
F/L = (4π × 10^(-7) T·m/A) * (6.00 A * 8.00 A * 0.14 m) / 0
F/L = 0 N/m
Therefore, the force per length exerted by I2 on I1 is 0 N/m. Again, due to the same-direction currents, the wires do not exert any net force on each other along their length.
Direction: Since the currents in both wires are in the same direction, the magnetic fields created by each wire at the location of the other wire will have the same direction. The forces exerted by the wires on each other will also be attractive, as the currents are flowing in the same direction.