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Find the magnetic induction due to a long straight conductor using Ampere's circuital law. |
Answer» Solution :Consider a straight conductor of infinite length carrying current I and the DIRECTION of magnetic field lines is shown in Figure. Since the WIRE is geometricallycylindrical in shape the wire is geometrically cylindrical in shapeand symmetrical about its axis, we construct an Amperian loop in the form of a circular shape at a distance r from the centre of the conductor as shown in Figure. From the AMPERE's law , we get `underset(C)ointvecB.vec(dl)= mu_(0)I` where `vec(dl) ` is the line element along the amperian loop ( tangent to the circular loop ). Hence, the angle between magnetic field vector and line element is zero . Therefore, `underset(C)ointBdl = mu_(0) I ` where I is the current enclosed by the Amperianloop. Due to the symmetry, the magnitude of the magnetic field is uniform over the Amperian loop, we can take B out of the INTEGRATION. `Bunderset(C)ointvec(dl) = mu_(0)I ` For a circular loop, the circumference is `2 pi r`, which implies, `Bunderset(0)overset(2 pi r)ointdl = mu_(0)I ` `vecB.2pir = mu_(0)I` `B = (mu_(0)I)/(2 pi r) ` In vector form, the magnetic field is `vecB = (mu_(0)I)/(2 pi r) hatn`where `hatn` is the unit vector along the tangent to the Amperian loop . |
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