A 30.0 cm diameter loop is rotated in a uniform electric field until the position of maximum electric flux is found. The flux in this position is 5.60 multiplied by 105 N·m2/C. What is the magnitude of the electric field?
MN/C
To find the magnitude of the electric field, we can use the formula:
Electric Flux = Electric Field * Area
Given in the problem:
Electric Flux = 5.60 * 10^5 N·m^2/C
Diameter of the loop = 30.0 cm
First, let's convert the diameter of the loop to its radius:
Radius (r) = Diameter (d) / 2 = 30.0 cm / 2 = 15.0 cm = 0.15 m
Next, let's calculate the area of the loop:
Area (A) = π * r^2 = π * (0.15 m)^2 = 0.0707 m^2
Now, we can rearrange the formula to solve for the electric field:
Electric Field = Electric Flux / Area
Electric Field = (5.60 * 10^5 N·m^2/C) / (0.0707 m^2)
Electric Field = 7.92 * 10^6 N/C
Therefore, the magnitude of the electric field is 7.92 * 10^6 N/C.