Question

Using the law of Biot and Savart, prove that the magnetic field generated by a loop of radius a carrying a current I at a point on the axis of the loop at a distance from the centre is В 2(12 + а?)3/2 Hint: Consider a generic element of length dl. Use the law of Biot and Savart for this element. How is the magnetic field generated by this element oriented? Consider an opposite element of same length. How is the magnetic field generated by this other element oriented? How is the sum of these two magnetic fields oriented? What is its magnitude? Now integrate over all possible pairs of element in the loop.

Answer 1

dBoost R = Jahr - - -> dBona #drona dBorsa Consider a circular loop of Radius a carrying Current I Let P e any point on the arris of the Coil. Consider a current element of Tength de According to Biot-Savart law dB= lo Idl 411 R² dB= ( loft) Idl R2 - Tatra) The Direction of dB will be perpendicular to -Both current element and R. which is making angle & with reatical axis Hence Vertical components will cancel out Due to anothy Symmetric current element and Horizontal component will add up. Bag dosis = dB sina

B= 1 Mo I adl UT (artal) 9/2 M. Ia (217a) UIT (x+2232 Integrating over length of loop - Cis conference of loop Mola 12 at (27a 2)/2 B = = Mo I a2 2(25+22) 312 to la B = 2(a2+92) 3/2 B= Mo I a2 21 22 +92) 312 Hence pooned
*The vertical component
dB Cosa)
Will cancel out with another diametrically present element on the loop and only horizontal component will add up.

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