A square singleturn wire loop phi 100 cm on a side is place

A square, single-turn wire loop phi = 1.00 cm on a side is placed inside a solenoid that has a circular cross section of radius r = 3.00cm, as shown in the end view of the figure below. The is 22.5 cm long and wound with 104 turns of wire. if the current in the solenoid is 3.05 A, what is that magnetic flues through the square loop? 1SC epsilon 8 T = m^2 if the current in the solenoid to in 3.25, what is the magnetic of the average inducted in their square loop?

Solution

Here, apply Ampère\'s circuital law -  
H * L = N * I

where,
H is the magnetic field intensity
L is solenoid length
N is the no. of winding turns
and, I is the current
So,

H * 0.225 = 104 * 3.05 [The data of your question is not very much clear. As I am reading, the current in the solenoid in part (a) is 3.05 A]

=> H = (104 * 3.05) / 0.225 = 1409.78

Again,
H * µ = B

Where,
B = Magnetic flux density
µ =1.256637061 ×10^6 [Magnetic permeability of the vacuum]
Then,
1409.78 * 1.256637061 x 10^6 =B

=> B = 1.77 x 10^-3 T

(a) Now, Magnetic flux = B * A

A = Inside area of the square loop = 0.01*0.01=10^-4 m^2

Therefore, magnetic flux crossing inside area of the square = 1.77 x 10^-3 x 10^-4 = 1.77 x 10^-7 T.m^2

(b) Emf in the square = N * d/dt(magnetic flux)
and, N = turns of square loop =1

Therefore, emf = 1 * ( 1.77x10^-7)/ (3.15-0) [ Again you are requested to varify the value of t, as this is not readable. I am considering it as 3.15 s]

=> emf = 5.62 x 10^-8 V