## anonymous one year ago 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? A. 1.0 * 10^-3 T B. 2.0 * 10^-3 T C. 8.0 * 10^-3 T D. 16 * 10^-3 T image drawn inside!!

1. anonymous

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2. Michele_Laino

the magnitude of the magnetic field inside a solenoid , is given by the subsequent formula: $\Large B = {\mu _0}\frac{{NI}}{L}$ where I is the current which flows along the solenoid, L is the xial length of the solenoid, N are the turns of our solenoid , and as usual: $\Large {\mu _0} = 4\pi {10^{ - 6}}Henry/meter$

3. anonymous

ok! what does that translate into for this problem? :/

4. Michele_Laino

so we can write: $\Large {B_1} = {\mu _0}\frac{{N{I_1}}}{L},\quad {B_2} = {\mu _0}\frac{{N{I_2}}}{L}$

5. anonymous

ok! what do we substitute in? :O

6. Michele_Laino

dividing side by side, we get: $\Large \frac{{{B_1}}}{{{B_2}}} = \frac{{{I_1}}}{{{I_2}}}$

7. Michele_Laino

now we have: I_1=3.20 A B_1=4.0 × 10^-3 T I_2=6.40 A so we can solve that equation for B_2, and we get: $\Large {B_2} = \frac{{{I_2}}}{{{I_1}}}{B_1} = \frac{{6.4}}{{3.2}} \times 4 \times {10^{ - 3}} = ...$

8. anonymous

and we get 0.008?

9. Michele_Laino

that's right!

10. anonymous

so would that mean choice A is the solution?

11. Michele_Laino

hint: 0.008 = 8*10^-3

12. anonymous

ohh okay! so choice C! thank you:)

13. Michele_Laino

:)