The worth of the line integral roughly the closed route in thefigure(Figure 1) is 1.96×10-5 Tm . What is the magnitudeof I3? The concepts compelled to deal with the offered trouble are Ampere’s circuital regulation, and right hand also rule.

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At first, use appropriate hand dominion to recognize the direction of unknown current enclosed by the loop. Then, recognize the net existing enclosed by the loop making use of direction of each existing inside the loop. Finally, determine the magnitude of unwell-known existing by using Ampere’s circuital law.

According to ideal hand preeminence, if the thumb of ideal hand also is alengthy the direction of current in a wire then the curl of continuing to be numbers offers the direction of magnetic field roughly the wire.

Ampere’s circuital regulation claims that the line integral of magnetic area B along a closed course is equal to μ0mu _0μ0​ times the current IencI_ mencIenc​enclosed by the closed route.

∮B⃗⋅dl⃗=μ0Iencoint vec B cdot dvec l = mu _0I_ menc ∮B⋅dl=μ0​Ienc​

Here, μ0mu _0μ0​is the permeability of free space.

The direction of unwell-known current I3I_3I3​is right into the web page bereason the direction of magnetic field is in clockwise direction.

The net present enclosed by the closed path is,

Ienc=I3−I2I_ menc = I_3 - I_2Ienc​=I3​−I2​

The Ampere’s circuital law is offered by adhering to expression.

∮B⃗⋅dl⃗=μ0Iencoint vec B cdot dvec l = mu _0I_ menc ∮B⋅dl=μ0​Ienc​

Substitute 1.96×10−5T⋅m1.96 imes 10^ - 5, mT cdot mm1.96×10−5T⋅mfor ∮B⃗.dl⃗,oint vec B.dvec l, ∮B.dl,4π×107T.m/A4pi imes 10^7, mT m.m/ mA4π×107T.m/Afor μ0mu _0μ0​, and I3−I2I_3 - I_2I3​−I2​for IencI_ mencIenc​in the over equation to settle for I3.I_3.I3​.

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1.96×10−5T⋅m=(4π×10−7T.m/A)(I3−I2)I3=1.96×10−5T⋅m(4π×10−7T.m/A)+I2=15.6A+I2eginarrayc\1.96 imes 10^ - 5, mT cdot mm = left( 4pi imes 10^ - 7, mT m.m/A ight)left( I_3 - I_2 ight)\\I_3 = frac1.96 imes 10^ - 5, mT cdot mmleft( 4pi imes 10^ - 7, mT m.m/A ight) + I_2\\ = 15.6, mA + I_2\endarray1.96×10−5T⋅m=(4π×10−7T.m/A)(I3​−I2​)I3​=(4π×10−7T.m/A)1.96×10−5T⋅m​+I2​=15.6A+I2​​

Substitute 12 A for I2I_2I2​in the above equation.

I3=15.6A+12A=27.6Aeginarrayc\I_3 = 15.6, mA + 12, mA\\ m = 27 m.6, mA\endarrayI3​=15.6A+12A=27.6A​