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?
5 answers
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?
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.
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.
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.
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.