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(a) Using Biot-Savart's law, derive the expression for the magnetic field in the vector form at a point on the axis of a circular current loop. (b) What does a toroid consist of? Find out the expression for the magnetic field inside a toroid for N turns of the coil having the average radius r and carrying a current I. Show the magnetic field in the open space inside and exterior to the toroid is Zero. |
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Answer» Solution :(a) Magnetic field on the axis of a circular LOOP I `rarr` Current , R `rarr` Radii X axis `rarr` Axis : x`rarr` Distance of OP DL `rarr` Conducting element of the loop According to Biot - Savart's Law, the magnetic field at P is `dB = (mu_(0)I |dl xx r|)/(4 pi r^(3))` `r^(2) = x^(2) + R^(2)` `|dl xx r| = r dl""(because "they are perpendicular")` `dB = (mu_(0))/(4 pi).(I dl)/((x^(2)+R^(2)))` dB has two components - dBx and `dB_(bot).dB_(bot)` is cancelled out and only the x-component remains. `therefore""dB_(x)=dB cos theta rArr cos theta=(R)/((x^(2)+R^(2))^(1//2))` `therefore""dB_(x)=(mu_(0)Idl)/(4PI)(R)/((x^(2)+R^(2))^(3//2))` Summation of dl over the loop is given by `2 pi R`. `therefore"'B=Bx hat(i)=(mu_(0)IR^(2))/(2(x^(2)+R^(2))^(3//2))hat(i)`.  (b) Toroid is a hollow circular ring on which a large number of turns of a wire are closely wound. Figure shows a sectional view of the toroid. The direction of the magneitc field inside a clockwise as per the right-hand thumb RULE for circular loops. Three circular Amperian loops 1, 2 and 3 are shown by dashed lines.  By symmetry, the magnetic field should be tangential to each of them and constant in magnitude for a given loop. Let the magnetic field inside the toroid be B. We shall now consider the magnetic field at S. Once again we employ Ampere's Law in the form of `oint vec(B).vec(dI)=mu_(0)I` `BL = mu_(0)NI` where, L is the length of the loop for which B is tangential, I be the current enclosed by the loop and N be the number of turns. We FIND, `""L = 2 pi r` The current enclose I is NI. `B (2 pi r) = mu_(0)NI`, therefore, `""B = (mu_(0)NI)/(2 pi r)` Open space inside the toroid encloses no current thus, I = O Hence, `""B = O` Open space exterior to the toroid - Each turn of current wire is cut twice by the loop 3. Thus, the current coming out of the plane of the paper is cancelled exactly by the current going into it . Thus, `I = O,""and""B = O` |
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