Answer:
The current in the coil is 4.086 A
Step-by-step explanation:
Given;
radius of the circular coil, R = 2.5 cm = 0.025 m
number of turns of the circular coil, N = 740 turns
magnetic field at the center of the coil, B = 0.076 T
The magnetic field at the center of the coil is given by;
![B = (N\mu_o I)/(2R)](https://img.qammunity.org/2021/formulas/physics/college/m3duvfx0opmey98wvrg1zmwf2haxz0gmoy.png)
where;
μ₀ is permeability of free space = 4 x 10⁻⁷ m/A
I is the current in the coil
R is radius of the coil
N is the number of turns of the coil
The current in the circular coil is given by
![B = (N\mu_o I)/(2R) \\\\I = (2BR)/(N\mu_o) \\\\I =(2*0.076*0.025)/(740*4\pi*10^(-7)) \\\\I = 4.086 \ A](https://img.qammunity.org/2021/formulas/physics/college/znz9rzfbhiu9p3xnw3kmnirfujmrqjwe5e.png)
Therefore, the current in the coil is 4.086 A