A 40.0cm diameter loop is turned to the position where the largest flux is handled in a uniform electric field. In this position, the flux is obtained as 5.20 × 10 ^ 5 N.m ^ 2 / c. Find the twist of the electric field.
To find the twist of the electric field, we need to use the formula for flux through a loop:
Φ = E * A * cos(θ)
Where:
Φ is the flux through the loop,
E is the electric field strength,
A is the area of the loop, and
θ is the angle between the electric field and the normal to the loop.
In this case, we are given the diameter of the loop, so we can calculate the area of the loop:
A = π * (r^2)
Where r is the radius of the loop, which is half the diameter. So,
r = 40.0 cm / 2 = 20.0 cm = 0.20 m
A = π * (0.20 m)^2 = 0.1257 m^2
Now, we can rearrange the formula for flux to solve for the electric field strength (E):
E = Φ / (A * cos(θ))
We are given the flux (Φ) as 5.20 × 10^5 N.m^2/C. However, we still need to find the angle (θ) between the electric field and the normal to the loop. Since the problem states that the loop is turned to the position where the largest flux is obtained, we know that the electric field and the normal to the loop are parallel. In this case, the angle between them will be 0°, or cos(θ) = cos(0°) = 1.
Now, we can calculate the electric field strength (E):
E = (5.20 × 10^5 N.m^2/C) / (0.1257 m^2 * 1)
E ≈ 4.14 × 10^6 N/C
So, the twist of the electric field is approximately 4.14 × 10^6 N/C.