advertisement

ELEG3203 2015-2016 First Term Assignment 2 1155047511 Li Man Hin 2. Divide the wire into three parts: two long straight wire and one semicircular coil. Using right hand grip rule, all three parts provides an out-of-paper magnetic flux density at point P. By symmetry, the two long wire can be considered as one long wire extending to both infinities. By Biot-Savart Law, Z L Z b dz 0 L µ0 I µ0 I µ0 I lim lim √ = aφ = aφ B = dB = aφ 02 2 3/2 2 2 4π L→∞ −L (z + b ) 2πb L→∞ L + b 2πb The magnetic flux density due to the half-coil is given by substituting z = 0 into (6-38). Thus we have µ0 I Bhc = 4b Total magnetic flux density is µ0 I µ0 I + â B= 2πb 4b where the direction of â is given by right-hand grip rule, i,e, out of paper. 1 3. Suppose we divide each wire into n each segments. Define dl = w n Consider an arbitrary point P (x, y, z) inside the coil where − w w < x, y < 2 2 Calculate the distance between P and By Biot-Savart Law, each line segment contributes a magnetic flux density dBi = where µ0 I dl × sin θi 4π Ri2 q Ri = ∆x2i + ∆yi2 + ∆zi2 and p 2 2 ∆yi + ∆zi , for two horizontal wires i sin θi = p R 2 ∆x + ∆zi2 i , for two vertical wires Ri The direction dBi is determined by right hand grip rule. Decompose it into the x, y, z direction, then sum up the dBix , dBiy , dBiz separately for : (a) x = −w/2, y = −w/2 to w/2 (b) x = w/2, y = −w/2 to w/2 (c) y = −w/2, x = −w/2 to w/2 (d) y = w/2, y = −w/2 to w/2 2