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A long straight metal rod has a very long hole of radius 'a' drilled parallel to its axis as shown in the figure.If the rod carries current i. find the magnetic field on the axis ofthe hole given C is the centre of the hole and OC = c.x |
Answer» Solution : In the rod current density `j = (i)/(pi(B^2 - a^2))` On the hole AXIS, only the larger rod contributes magnetic field. Imagine an amperean loop of radius C and apply Ampere.s law. `B_("Total") = UNDERSET(underset("solid")"complete")(vec(B_1)) + underset("cavity")(vec(B_2)) = (mu_0 j pi c^2)/(2 pi c) + O` `= (mu_0ic)/(2pi(b^2-a^2)) = (mu_0)/(2pi) (ic)/((b^2 - a^2))` |
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