Answer
given,
Radius of the circular current = 50 mm
Number of turn = M = 2 turn/mm
magnetic field at the center of the loop
[tex]B = \dfrac{\mu_0 I}{2R}[/tex]
magnetic field at the center of solenoid
[tex]B' = \mu_o n I[/tex]
now, ratio
=[tex]\dfrac{B'}{B} = \dfrac{ \mu_o n I}{\dfrac{\mu_0 I}{2R}}[/tex]
=[tex]\dfrac{B'}{B} =2 n R[/tex]
=[tex]\dfrac{B'}{B} =2\times 2 \times 50[/tex]
B' = 200 B
magnetic field inside the solenoid is 200 times the magnetic field of loop.
