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Hello students we are given two concentric cylinders such that their length is out of the paper that is along the z direction.
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R a and rb are the radii of the two cylinders.
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The inner cylinder is a solid cylinder and the current through it is 4 .5 ampers out of the page and the outer cylinder is a hollow cylinder and the current through it is 7 .5 ampiers into the page.
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We need to find out the magnetic field at two points a and c, which are along the y -axis.
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For doing so, we assume two imperial loops with radii equal to a and c as shown.
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We will be applying the amperous circuital law to these two empirian loops to find the magnetic field at point a and point c.
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First, to find the magnetic field at point a, we use the amperous circuital law given as b .dl equals to mu0 into i, where i is the current enclosed by the loop.
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Or this can be written as magnetic field at point a into 2 pi r a equals to mu zero into i enclosed which gives us magnetic field at a will be equal to mu zero into i enclosed upon 2 pi r into a let's say this is equation one now to find i enclosed we use the enclosed current will be equal to current density that is current flowing per unit area multiplied by the cross -sectional area of the amperian loop or this is equal to 4 .5 upon pi into 0 .08 square into pi into 0 .04 square.
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Solving this we get the value of enclosed current is 1 .125 amper.
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Thus, substituting the values in equation 1 we get magnetic field at a will be equal to 4 into 3 .14 into 10 to the power minus 7 multiplied by 1 .125 upon 2 into 3 .14 into 0 .04 which gives us b a will be equal to 5 .625 into 10 to minus 6 tesla thus your answer is the magnitude of magnetic field at a is equal to 5 .62 into 10 to the bar minus 6 tesla...