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!!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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!!

Physics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

|dw:1434435467105:dw|
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: \[\Large {\mu _0} = 4\pi {10^{ - 6}}Henry/meter\]
ok! what does that translate into for this problem? :/

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

so we can write: \[\Large {B_1} = {\mu _0}\frac{{N{I_1}}}{L},\quad {B_2} = {\mu _0}\frac{{N{I_2}}}{L}\]
ok! what do we substitute in? :O
dividing side by side, we get: \[\Large \frac{{{B_1}}}{{{B_2}}} = \frac{{{I_1}}}{{{I_2}}}\]
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: \[\Large {B_2} = \frac{{{I_2}}}{{{I_1}}}{B_1} = \frac{{6.4}}{{3.2}} \times 4 \times {10^{ - 3}} = ...\]
and we get 0.008?
that's right!
so would that mean choice A is the solution?
hint: 0.008 = 8*10^-3
ohh okay! so choice C! thank you:)
:)

Not the answer you are looking for?

Search for more explanations.

Ask your own question