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2. N B(z) T 0 a plout) N (A) (с (B) Figure 18 Example 5. A circular loop of N turns and radius a in the xy plane carries a steady current I (see Figure 18A] Νμο 4 Motor L" ladd\-sinfi + cos d'i) x cua co di – asin d'ž +_k] L" + ) 47(22 +22)7|04 +0j + 2rak k_N Hola (a)Show that the magnetic field at a point P on the axis of the loop, at a distance z from the center is: B = k_Nula = kB(z). Solution: 2122+e) B Id5' x () N HOT (ado (z cos d'i + z sind) + ad¢ £] Νμο! (z2 + a)? kB(2) 2(22 + a)! (b) The force on a magnetic dipole jī in presence of a magnetic field is given by F = 76-8). Show that if i = jk is located at : above the circular loop then the force on the dipole is attractive and is given by F = - Suple: - 3 B(z) (note that below the circular 2(22+) loop the force is again attractive). Solution: Å B # = õlā. B) = (1 + i + GLE) - , Nuola? +k a Nuola? dr az az 2(22+) =-3UB(2) 2(-2+)! - a a öy ( z 222 +22)])) = = -23 MM/a> +

(e) Let ji = (-sin bi + cos 83) at z. Show that the torque on the needle is 7 = 1x B = B(2) (cos oi+sin 03) = Nola? P21212, see figure 9(B)]. Solution j1 x B = rl-sin 6i + cose) * B{z}k = B(z)(+sin 8j + cosi) = B(z) Nuola? 2(32 + a)? (d) [Helmholtz coil, see figure 9(C)Show that for two identical coils of radius a, separation distance a, and carrying the current I in the same direction, the magnetic field along the symme try axis can be written as: B(s) = k - [*** pe ] - BO) – konge Solution This follows from part(a) by a translation: Bus – Ētz - +B(+9) where B = 1 Nosła? 2(22+) (e) Show that the field is quite uniform about == 0 by evaluating the first few derivatives of magnetic field at := 0; = 0 i.e. the Taylor-Melauren expansion about 20 then gives B BO 2 ав 1! az 2 B 2! 87 PB ++ 3! 823 OB 4! + +0(2) RANE -0.824304 Viele -0.1920 8140 C+0(3) av 125 Solution: This is straight forward but tedious: oglu – 1Naro( Taufe 13" (2+4*3*" (-3) 15() Poffc = NI2*10 (2462=39** (2+12+1)3* (2+12+0)7*** (+6+)*)* (-))*(7(82 (3+ 106(5+) = Nav (642-37* * (++(3-1)*)*** (2+(-9) (2+(3+)")" (+(4+)*)** as -

(f)If the direction of the current I on the loop at z = -a/2 is reversed and a dipole j = pk is placed at the origin then: B(z) = 20 + +237 BO) = 0 ; . ) -Νμο/α? Nola? 2([=+9P+c)2(12-1P+c?) - OBE= 3 Nasle?jz+ Ngota :- [2(12+3P+22) 212-3P+a?) (True, False) 6 SNTA 68NT 5 1250 0 Summary Here is the summary on the force on a dipole pointing in the z direction (i)for currents in the same direction B(z) = Nola Nola |--${N+vola (2-193** (+) 573 2012 Hyltos and videos) - $INA 40 () P = 240 = $1N*410 (Content 3(3-3) (22+(3-1))** (23+(3+2) *) *** - for currents in the opposite direction: -Nuolaa + - சா B(z)= Nuolo 2(15+9P+a)} *21-9P+2)! OBW) = 36 pola ista Nuolas-1 2(1=+P+ 23-3P+) F = 1080) = 3 Nuola?[s+JĮ polas-1 2(12+9P+c) 2[=-3P+c) In figure 19 the force [-(--B)) and the potential energy U =-ji- B is plotted as a function of z.