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Eight wires cut the page perpendicular t...

Eight wires cut the page perpendicular to the points shown. Each wire carries current `i_(0)`. Odd currents are out of the page and even current into the page. Find the line integral `ointvec(B).vec(dl)` along the loop.

Text Solution

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According to ampere.s circuital law
`oint vecB.vecdl = mu_0 i_("enclosed") , oint vecB.vecdl = mu_0 [ i_8- i_s + i_2 + i_4]`

Since , all the wires carry same current of `i_0` , we have `oint vecB.vecdl = 2mu_0 i_0`
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Knowledge Check

  • Eight wires cut the page perpendicularly at the points shown. Each wire carries current i. Odd currents are out of the page and even currents into the page. The line integral int vecB. vec dl -along the loop is

    A
    `mu_(0)i_(0)`
    B
    `2mu_(0)i_(0)`
    C
    0
    D
    `3mu_(0)i_(0)`

  • Eight wires cut the page perpendicularly at the points shown in figure . A wire labeled with the integer k(k=1.2…..80 carries the current ki, where i=2A . For those wires with odd k, the current is out of the page . For those with even k, it is into the page. the value of ointvecB.dvecs=0 along the closed path indicated and in the direaction shown.

    A
    `10mu_(0)`
    B
    `5mu_(0)`
    C
    `15mu_(0)`
    D
    `20mu_(0)`

  • Eight wires cut the page perpendicularly at the points show in figure. A wire labeled with the integer k (k=1.2,…….8) bears the current ki_(0) . For those with odd k, the current flows up out of the page, for those with even k it flows down into the page. The value of ointvecB.dvecr along the close path (as shown in the figure) in the direction indicated by the arrow is

    A
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    B
    `-10mu_(0)i_(0)`
    C
    `-4mu_(0)i_(0)`
    D
    `4mu_(0)i_(0)`

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