A current of 3.2 A in a solenoid produces a magnetic field of 4 x 10^-3 T. Current is changed to 6.4 A . What's the magnitude of magnetic field after change?

Question

anonymous · Answer

This website could help http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magcur.html

anonymous · Answer

okay I already saw that but didn't quite understand it so that's why I came on here

anonymous · Answer

basically you can use the equation from that site, then plug in the number, since it is the same solenoid, the number of turns and radius is the same so those 2 variables can safely be ignored, only I (current) is changing.

anonymous · Answer

magnitude here is B - the generated magnetic field

anonymous · Answer

ok thanks

anonymous · Answer

@zaphod @whpalmer4

anonymous · Answer

I think this will work i am not sure
$$B = \frac{\mu_{o} \times I}{2 \pi X }$$
You hve been given the value of B and I for the first instant which i assume would give the constants 

$$B = k \times I$$

You might want to find the value of k, using the first valyes of B and I by plugging them into the equation right above

Then using that value of k u can find the new B for the I they hve given

$$B_{1} = k \times I_{1}$$

You can ffind B1, since they are asking the change u cn subtract (b1- b) = change in magnetic flux

anonymous · Answer

sorry, u just need to find B1 not the change, so u dont hve to subtract :)

whpalmer4 · Answer

$B$ varies directly with $I$, so doubling the current from 3.2 A to 6.4 A should also double the magnetic field.