Get 5 free video unlocks on our app with code GOMOBILE. 5) Two loops are placed near current-carrying wires as shown in Case and Case 2 below: In both cases, the direction of the current in the two wires is opposite to each other:

From Ohm’s law applied between 1 and 7 via 1487 (say)(b) Between 1 and 2 from the loop 14321 , $I_{1} R=2 I_{2} R+I_{3} R$ or, $I_{1}=I_{3}+2 I_{2}$From the loop 48734 , $\left(I_{2}-I_{3}\right) R+2\left(I_{2}-I_{3}\right) R+\left(I_{2}-I_{3}\right) R=I_{3} R$or, $4\left(I_{2}-I_{3}\right)=I_{3}$ or $I_{3}=\frac{4}{5} I_{2}$so $I_{1}=\frac{14}{5} I_{2}$Then, $\left(I_{1}+2 I_{2}\right) R_{e q}=\frac{24}{5} I_{2} R_{\mathrm{eq}}=I_{1} R=\frac{14}{5} I_{2} R$or $R_{e q}=\frac{7}{12} R$(c) Between 1 and 3From the loop 15621$$\begin{aligned}&\qquad I_{2} R=I_{1} R+\frac{I_{1}}{2} R \text { or, } I_{2}=3 \frac{I_{1}}{2} \\&\text { Then, } \begin{array}{c}\left(I_{1}+2 I_{2}\right) R_{e q}=4 I_{1} R_{e q} \\& =I_{2} R+I_{2} R=3 I_{1} R \\\text { Hence, } \quad R_{e q}=\frac{3}{4} R\end{array}\end{aligned}$$. In the figure, two parallel wires carry currents of magnitude I

The induced current in the loop is decreasing with time. Two identical current carrying coaxial loops, carry current $I$ in an opposite sense

Ampere’s Law is similar to Gauss’ Law, as it allows us to (analytically) determine the magnetic field that is produced by an electric current in configurations that have a high degree of symmetry. where the integral on the left is a “path integral”, similar to how we calculate the work done by a force over a particular path

\(I^{enc}\) is the net current that crosses the surface that is defined by the closed path, often called the “current enclosed” by the path. This is different from Gauss’ Law, where the integral is over a closed surface (not a closed path, as it is here)

We apply Ampere’s Law in much the same way as we apply Gauss’ Law.. – Choose a closed path over which to calculate the circulation of the magnetic field (see below for how to choose the path)

The gray circle in the center represents a cross section of a wire carrying current coming out of the computer screen. The current is uniformly distributed throughout the wire (position is given in centimeters and magnetic field strength is given in millitesla)

Begin with the Amperian loop with a radius larger than the radius of the wire.. You will use Ampere’s law to find the total current in the wire:

Pick a point on the Amperian loop and draw both the direction of the magnetic field at that point and the direction of dl (tangent to the path).. – The magnetic field and dl should be parallel to each other

Ampere’s Law We know for an infinite current carrying wire. Ampere’s Law Any closed loop Current enclosed by that closed loop

For which loop is ∫B·dl greater? Case 1 Case 2 The integral is the same for both. Checkpoint 1B Two loops are placed near identical current carrying wires as shown in Case 1 and Case 2 below

Checkpoint 1C Two loops are placed near current carrying wires as shown in Case 1 and Case 2 below. In both cases the direction of the current in the two wires are opposite to each other

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Ampere’s Law We know for an infinite current carrying wire But what is 2pR? Circumference of circle!. Ampere’s Law Any closed loop Current enclosed by that closed loop

• For which loop is ∫B·dl greater? • Case 1 • Case 2 • The integral is the same for both. Checkpoint 1B Two loops are placed near identical current carrying wires as shown in Case 1 and Case 2 below

Checkpoint 1C Two loops are placed near current carrying wires as shown in Case 1 and Case 2 below. In both cases the direction of the current in the two wires are opposite to each other