A current of 3.20 A in a solenoid produces a magnetic field of 4.0 × 10^-3 T. The current is changed to 6.40 A. What is the magnitude of the magnetic field after the change?

Question

anonymous · Answer

attachment

anonymous · Answer

Thank you for coming to help me again

irishboy123 · Answer

for a solenoid, magnetic field strength  $B = \mu n I$ =  permeability $ \mu \times$  turn density $n \times$ current $I$.

all you have done is double the current, so the magnetic field, B, should also do what?

anonymous · Answer

16 × 10^-3 T
8.0 × 10^-3 T <----- would this be the answer?
2.0 × 10^-3 T
1.0 × 10^-3 T

irishboy123 · Answer

yay!!