A rectangular coil of dimensions 20cm by 15cm lies with its plane parallel to a magnetic field of 0.5W/m^2. The coil, carrying a current of 10A experiences a torque of 4.5Nm in the field. How many loops has the coil?
N = 30 loop
A rectangular coil of 25 loops is suspended in a field of 0.20 Wb/m?. The plane of the coil is parallel to the
direction of the field. The dimensions of the coil are 15 cm verendicular to the field lines and 12 cm paralle. to them. What is the current in the coil if there is a torque of 5.4 N-m acting on it?
We can use the formula for the torque experienced by a rectangular coil in a magnetic field:
τ = NABsinθ
where τ is the torque, N is the number of turns in the coil, A is the area of the coil, B is the magnetic field strength, and θ is the angle between the normal to the coil and the direction of the magnetic field.
We can rearrange this equation to solve for the current I:
I = τ/(NABsinθ)
We are given N = 25, A = 0.15 x 0.12 = 0.018 m^2, B = 0.20 Wb/m^2, and θ = 0° (since the plane of the coil is parallel to the field lines). We are also given τ = 5.4 N-m.
Substituting these values, we get:
I = 5.4/(25 x 0.018 x 0.20 x sin0°) = 6 A
Therefore, the current in the coil is 6 A.
A closely wound, flat, circular coil of 25 turns of wire has a diameter of 10 em and carries a current of 4.0 A
Determine the value of B at its center.
We can use the formula for the magnetic field at the center of a circular coil:
B = μ0IN/2R
where B is the magnetic field strength, μ0 is the permeability of free space (4π x 10^-7 Tm/A), I is the current in the coil, N is the number of turns in the coil, and R is the radius of the coil (which is equal to half the diameter).
We are given I = 4.0 A, N = 25, and the diameter of the coil is 10 cm, so the radius is 5 cm.
Substituting these values, we get:
B = (4π x 10^-7 Tm/A) x (4.0 A) x (25) / (2 x 0.05 m)
Simplifying this expression, we get:
B ≈ 0.010 T
Therefore, the magnetic field strength at the center of the coil is approximately 0.010 T.
To find the number of loops in the coil, let's first understand the concept of torque in a magnetic field:
When a current-carrying coil is placed in a magnetic field, it experiences a torque, which tends to align the coil with the magnetic field. The torque can be calculated using the formula:
Torque = N * B * A * sin(theta)
Where:
N is the number of loops
B is the magnetic field strength
A is the area of the coil
theta is the angle between the magnetic field and the normal to the coil's plane
In this case, we know the torque, magnetic field strength, area of the coil, and current. We are given that the torque is 4.5 Nm, the magnetic field strength is 0.5 W/m^2, and the dimensions of the coil are 20 cm by 15 cm.
To get the area of the coil, convert the dimensions from centimeters to meters:
Length = 20 cm = 20/100 m = 0.2 m
Width = 15 cm = 15/100 m = 0.15 m
Area = Length * Width = 0.2 m * 0.15 m = 0.03 m^2
Now, let's rearrange the torque formula to solve for the number of loops (N):
N = Torque / (B * A * sin(theta))
Since the coil lies parallel to the magnetic field, the angle (theta) between the magnetic field and the normal to the coil's plane is 90 degrees, and sin(90) is 1. Therefore, we can simplify the equation:
N = Torque / (B * A)
Substituting the given values:
N = 4.5 Nm / (0.5 W/m^2 * 0.03 m^2)
N = 4.5 Nm / (0.015 W)
N = 300 loops
Therefore, the coil consists of 300 loops.