You have a wire of length L = 2.1 m for making a square coil of a dc motor. The current in the coil is I = 1.8 A, and the magnetic field of the motor has a magnitude of B = 0.32 T. Find the maximum torque exerted on the coil when the wire is used to make (a) a single-turn square coil and (b) a two-turn square coil.
CORRECTION TO ELENA's ANSWER:
M2=2*I*L^2*B/(8^2)
A₁=(L/4)²
M₁=p₁•B=I•A₁•B=I•L²•B/16=…
A₂=(L/16)²
M₂=2p₂•B=2I•A₂•B=2I•L²•B/256=…
M2 = M1/2
To find the maximum torque exerted on the coil, we can use the formula for the torque on a current-carrying coil in a magnetic field:
τ = N * I * A * B * sin(θ)
Where:
τ is the torque
N is the number of turns in the coil
I is the current in the coil
A is the area of the coil
B is the magnetic field
θ is the angle between the magnetic field and the normal to the coil
Let's solve for each case separately:
(a) Single-Turn Square Coil:
In this case, N = 1 (single turn), and the area of the coil (A) can be calculated by squaring the length of one side of the coil (L):
A = L^2
Substituting these values into the torque formula:
τ = 1 * I * L^2 * B * sin(θ)
(b) Two-Turn Square Coil:
In this case, N = 2 (two turns), and the area of the coil (A) can still be calculated by squaring the length of one side of the coil (L):
A = L^2
Substituting these values into the torque formula:
τ = 2 * I * L^2 * B * sin(θ)
Note: In part (b), we multiply by 2 because there are two turns in the coil.
To calculate the maximum torque in each case, we need to know the value of the angle θ. Without this information, we cannot determine the precise value of the torque.