**Outside** the wires, the two magnetic areas from the 2 curleas are in **opposite** directions so they will tend to cancel each various other and the resulting or total magnetic field will certainly be **small**. Q4 Exsimple why two parallel wires moving currents in opposite directions repel each various other. You might desire to look at the diagrams for Inquiry 3. In specific, think about this diagram,

or

The magnetic area of the left-hand also present causes a force on the right-hand also present that is a repulsion so we understand -- from Newton"s Third Law -- that the two curleas repel each various other. Q9 Describe the similarities between Ampere"s legislation in magnetism and Gauss"s law in electrostatics. Both associate an integral around a closed surconfront or line with something that goes with the surface or line of the integral. Gauss"s Law says the electrical flux through a closed surface is proportional to the charge within that surface.. Ampere"s Law states the line integral of B-ds approximately a closed route is proportional to the present that passes with that route. Q18 Will a nail be attracted to either pole of a magnet? Explain what is happening inside the nail once placed near a magnet. Yes, a nail will certainly be attracted to either pole of a magnet. The magnet causes the boundaries in between magnetic domains to relocate slightly, providing a net magnetic area to the nail.course If the nail is near the othermagnetic pole then various magnetic doprimary boundaries relocate. Q19 The north-seeking ppole of a magnet is attracted towards the geographical North Pole of Earth. Yet, choose poles reple. What is going on here? The magnetic pole near Earth"s geographic North Pole is what would otherwise be dubbed a south-seeking magnetic pole. 30.8A, A conductor is composed of a circular loop of radius R and also two directly, long sections, as in Figure P30.8. The wire lies in the plane of the paper and carries a existing I. Determine the magnitude and also direction of the magnetic area at the facility of the loop. We recognize exactly how to manage the magnetic field as a result of a directly wire; At a distance R from the wire, the magnetic area is And we recognize just how to manage the magnetic field at the center of a circular loop; At the center of the wire, the magnetic field is So our total magnetic field at the center of the circular loop and also distance R from the long, directly wire, is the sum of these, 30.22, A carefully wound, long solenoid of all at once length 30.0 cm (0.30 m) has a magnetic field of 4.00 x 10 - 4 T at its center created by a current of 1.00 A via its windings. How many type of transforms of wire are on the solenoid? The magnetic field inside a solenoid is offered by or n = B / I n = (4 x 10 - 4) / <(4 x 10 - 7)(1)> n = 318 turns/m n = N/L N = n L N = (318 turns/m)(0.30 m) N = 95 transforms 30.23, A superconducting solenoid is to geneprice a magnetic area of 10.0 T. (a) If the solenoid windings has 2000 turns/m, what is the compelled current? or I = B / n I = (10)/<(4 x 10 - 7)(2000)> I = 3979 A That is an massive current; yet then that is an massive magnetic field! 30.25, Some superconducting alloys at exceptionally low temperatres have the right to lug incredibly high currents. For instance Nb3Sn wire at 10 K can bring 1000 A and also maintain its superconductivity. Determine the maximum value of B achievable in a solenoid of length 25 cm (0.25 m) if 1000 turns of Nb3Sn wire are wrapped on the outside surchallenge. B = (4 x 10 - 7)(1000/0.25)(1000) B = 5 T 30.32, In Figure P30.32, both curleas are in the negative x direction. (a) Map out the magnetic area pattern in the yz-aircraft. Remember, this is a rough sketch! 30.40, Consider the hemispherical closed surchallenge in Figure P30.40. If the hemisphere is in a uniform magnetic area that renders and also angle theta with the vertical, calculate the magnetic flux through (a) the flat surconfront S1 and also = A B cos A = R2 = R2 B cos (b) the hemispherical surconfront S2. According to Gauss"s Law for magnetic flux, the total magnetic flux is zero. Due to the fact that we recognize the flux coming into the surconfront ( = R2 B cos ) that (or the negative) of that is likewise the flux coming out of the surface -- the flux through S2, = - R2 B cos 30.60, A lightning bolt may lug a current of 10,000 A for a short period of time. What is the resulting magnetic area 100 m from the botl? B = <(4 x 10 - 7)(10 000)>/<(2 ) (100)> B = 2 x 10 - 5 T 30.64 Two parallel conductors lug present in oppowebsite directions as showin in Figure P30.64. One conductor carries a present of 10A. Point A is at the midpoint in between the wires and suggest C is a distance d/2 to the ideal of the 10 A present.You are watching: Explain why two parallel wires carrying currents in opposite directions repel each other.

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If d = 18 cm = 0.18 m and also I is adusted so that the magnetic area at C is zero, discover (a) the worth of the existing I and (b) the value of the magnetic area at A. For Point C, Bappropriate = <(4 x 10 - 7)(10)>/<(2 )(0.09)> = 2.22 x 10 - 5 T Bleft = <(4 x 10 - 7)( I )>/<(2 )(0.27)> = 2.22 x 10 - 5 T I = (2.22 x 10 - 5) (0.27) / 2 x 10 - 7 I = 30 A For Point A, Bright = <(4 x 10 - 7)(10)>/<(2 )(0.09)> = 2.22 x 10 - 5 T Bleft = <(4 x 10 - 7 )( 30 )>/<(2 )(0.09)> = 6.66 x 10 - 5 T Using the Right Hand also Rule, we view that these two fields point in the same direction -- "up" or "out of the page" at allude A Btotal = Bright + Bleft = 8.88 x 10 - 5 T Synopsis Ch31 ToC Return to ToC (c) Doug Davis, 2002; all rights reserved