Question

Calculate the current which must pass through an ideal solenoid so that in its interior a magnetic field of 200 mT is generated. The solenoid has 800 turns (loops) of a wire of radius 4.0 cm.

Answer 1

$B l= \mu_0 N I,$

using this I=Bl/N(uo)

I=200*10^-3*4*!0^-2/(800*1.26*10^-6)

I=7.94 amperes

Answer 2

we know that

B = (u_o x N x I/2pi x a)

or I = (B x 2pi x a/u_o x N)

where I is current,B = 200 mT = 200 x 10^-3 T,a = 4.0 cm = 4.0 x 10^-2 m,u_o = 4pi x 10^-7 T.m/A and N = 800 turns