A rectangular coil 60mm × 40mm with 25 turns is positioned with its longer sides parallel to a long straight vertical wire carrying a current of 6.8A. The coil is placed so that its long sides and the vertical wire are in the same vertical plane, with the wire 40mm away from the nearest long side of the coil. A current of of 2.0A is passed round the coil.
(a). Calculate the force on each long side of the coil due to the magnetic field created by the vertical wire.
(b). Explain why the top and bottom sides of the coil experience equal and opposite forces.
(c). Calculate the overall force on the coil.
To solve this problem step-by-step, we'll break it down into several parts:
(a) Calculating the force on each long side of the coil due to the magnetic field created by the vertical wire:
1. Calculate the magnetic field created by the vertical wire at the position of the coil:
- The magnetic field created by a current-carrying wire at a distance r from the wire can be calculated using the formula:
B = (μ₀ * I) / (2πr)
where B is the magnetic field, I is the current, r is the distance from the wire, and μ₀ is the permeability of free space (4π × 10⁻⁷ T·m/A).
- Plugging in the values:
I = 6.8 A (current in the wire)
r = 40 mm = 0.04 m (distance from the wire)
B = (4π × 10⁻⁷ T·m/A * 6.8 A) / (2π * 0.04 m)
= 5.44 × 10⁻⁵ T
2. Calculate the magnetic flux through each long side of the coil:
- The magnetic flux through a coil can be calculated using the formula:
Φ = B * A * cos(θ)
where Φ is the magnetic flux, B is the magnetic field, A is the area of the coil, and θ is the angle between the magnetic field and the normal to the coil.
- In this case, the magnetic field and the normal to the coil are parallel, so the angle θ is 0 degrees.
- Plugging in the values:
B = 5.44 × 10⁻⁵ T (calculated earlier)
A = 60 mm × 40 mm = 0.06 m × 0.04 m = 0.0024 m²
Φ = (5.44 × 10⁻⁵ T) * (0.0024 m²) * cos(0°)
= 5.44 × 10⁻⁵ * 0.0024
= 1.3056 × 10⁻⁷ Wb (Weber)
3. Calculate the force on each long side of the coil:
- The force on a current-carrying wire in a magnetic field can be calculated using the formula:
F = Φ * I * sin(θ)
where F is the force, Φ is the magnetic flux, I is the current in the wire, and θ is the angle between the magnetic field and the current direction.
- In this case, the current in the wire and the magnetic field are parallel, so the angle θ is 0 degrees.
- Plugging in the values:
Φ = 1.3056 × 10⁻⁷ Wb (calculated earlier)
I = 2 A (current in the coil)
F = (1.3056 × 10⁻⁷ Wb) * (2 A) * sin(0°)
= 2.6112 × 10⁻⁷ N
Therefore, the force on each long side of the coil due to the magnetic field created by the vertical wire is approximately 2.6112 × 10⁻⁷ N.
(b) Explaining why the top and bottom sides of the coil experience equal and opposite forces:
- The top and bottom sides of the coil are parallel and have the same dimensions (60 mm × 40 mm) and the same number of turns (25). Therefore, the current passing through each side is the same.
- According to the right-hand rule, the direction of the force acting on a current-carrying wire in a magnetic field is perpendicular to both the current direction and the magnetic field direction. In this case, since the current in the coil passes in the same direction on both the top and bottom sides, the forces on these sides will be in opposite directions but equal in magnitude.
(c) Calculating the overall force on the coil:
- Since the forces on the top and bottom sides of the coil are equal and opposite, they cancel each other out.
- Therefore, the overall force on the coil is zero.
(a) To calculate the force on each long side of the coil due to the magnetic field created by the vertical wire, we need to use the formula for the magnetic force on a current-carrying wire.
The magnetic force on a wire can be calculated using the equation:
F = BIL
Where:
F is the force on the wire,
B is the magnetic field strength,
I is the current flowing through the wire,
L is the length of the wire.
In this case, we have a rectangular coil with dimensions 60mm × 40mm, so each long side has a length of 60mm. The coil has 25 turns, which means the total length of the wire is 60mm * 25 = 1500mm.
First, let's calculate the magnetic field strength at a distance of 40mm away from the vertical wire. We can use the formula for the magnetic field of a long straight wire:
B = (μ₀ * I) / (2 * π * r)
Where:
B is the magnetic field strength,
μ₀ is the permeability of free space (a constant),
I is the current flowing through the wire,
r is the distance from the wire.
The permeability of free space is approximately 4π × 10^(-7) T*m/A.
Substituting the values into the formula, we have:
B = (4π × 10^(-7) * 6.8) / (2 * π * 40mm)
Simplifying, we get:
B ≈ 2.72 * 10^(-6) T
Now we can calculate the force on each long side of the coil:
F = B * I * L
F = 2.72 * 10^(-6) T * 2.0 A * 1500mm
Note that 1500mm is used here since we are considering the entire length of the coil.
Simplifying and converting millimeters to meters:
F ≈ 8.16 * 10^(-3) N
Therefore, the force on each long side of the coil due to the magnetic field created by the vertical wire is approximately 8.16 millinewtons (mN).
(b) The top and bottom sides of the coil experience equal and opposite forces because they are on opposite sides of the vertical wire, and the direction of the magnetic field created by the wire is perpendicular to the plane of the coil. According to the right-hand rule, the magnetic force on a current-carrying wire is perpendicular to both the direction of the current and the direction of the magnetic field. Since the top and bottom sides of the coil carry currents in opposite directions, the forces they experience will be equal in magnitude but opposite in direction.
(c) To calculate the overall force on the coil, we can sum up the individual forces on each long side. Since the forces are equal and in opposite directions, they cancel each other out. Thus, the overall force on the coil will be zero.