Magnetostatics PDF

Application of Biot-Savart’s Law to Calculate Magnetic Field  The Biot-Savart Law The magnetic field of a steady line current is given by the Biot-Savart law:...

Application of Biot-Savart’s Law to Calculate Magnetic Field  The Biot-Savart Law The magnetic field of a steady line current is given by the Biot-Savart law: The integration is along the current path, in the direction of the flow; dl’ is an element of length along the wire, and r, as always, is the vector from the source to the point r. The constant μ0 is called the permeability of free space: The SI unit of B: In CGS unit of B: Newtons per ampere-meter (as required by the Lorentz force law), Gauss or Tesla (T) 1 tesla = 104 gauss. 1 T = 1 N/(A · m)  B In the case of an infinite wire, 1 = -π/2 and θ2 = π/2, so we obtain The vertical component of dB d B cos The horizontal component of dB d B sin dl’ r = dl’ sin (/2)= dl’ (a) First calculate the magnetic field due to the one side of the current carrying loop (ex. B1) B  you can use the above equation, with s = R, and set θ2 = – 1 = π/4  B1 = (0I/4R)[Sin(π/4) – Sin(– π/4 )] =  2 0I/4R  B = B1 + B2 + B3 + B4 (due to all four sides of the square loop) =  2 0I/4R (b) Similar to the part (a), set s = R, and set θ2 = – 1 = π/n in the Eq.37, and calculate the magnetic field due to the one side of the current carrying loop (ex. B1) B  B1 = (0I/4R)[Sin(π/n) – Sin(– π/n )] = (0I/2R)Sin (π/n)  B = B1 + B2 + B3 + …. Bn (due to all n sides of the n-sided polygon loop) = (n0I/2R)Sin (π/n) (c) Similar to the part (b), but to imagine polygon to be like a circle one has to consider, n  , that means,  is very small, therefore, sin    B set s = R, and set Sinθ2 = θ2= π/n and Sin1 = θ1=–π/n in the Eq.37, and calculate the magnetic field due to the one side of the current carrying loop (ex. B1)  B1 = (0I/4R)[Sin(π/n) – Sin(– π/n )] = (0I/4R) (π/n + π/n) = 0I/2nR  B = B1 + B2 + B3 + …. Bn (due to all n sides of the n-sided polygon loop) B = n0I/2nR = 0I/2R  The Divergence & Curl of Magnetic Field For the case of an infinitely long straight wire carrying a steady (constant) line current, the macroscopic magnetic field associated with this system is given by:  This clearly signifies that the magnetic field vector, B has circulation or vorticity associated with it.  Means  Compared to what we see in electrostatics, the electric field vector Application of Ampère’s Law to Calculate Magnetic Field Example 9. Find the magnetic field of a very long solenoid, consisting of n closely wound turns per unit length on a cylinder of radius R, each carrying a steady current I (Fig. 34). Magnetic field inside a solenoid is uniform – analogous to capacitor which produces uniform electric fields  2 0I/4R  Q. 4 Part-2 Electric field intensity at the position, O’ (5 cm above the origin, O) due to the loop, carrying uniform charge density. z (i) 1.81102 N/C O’ (ii) 1.8110-8 N/C 5 cm (iii) 1.81102 Nm2/C Charge density, (iv) 1.81108 N/C  = 5 C/cm Ans = (iv) O 2 cm y x  Q. 5 Part-2 Magnetic field at the center (O) of this current (10 Amp) carrying structure of equal side length is: (i) 0.173 Gauss (ii) 8.6610-5 Gauss O (iii) 1.7310-4 T (iv) 0.43 Gauss 4 cm Ans = (iii)  Q. 6 Part-2 The value of the magnetic field intensity at the position “O” due to the current carrying wire is given as: (i) 7.8510-5 T O 2 cm 5 Amp (ii) 1.57 Gauss (iii) 6.2510-4 T (iv) 1.2510-4 T Ans = (i)