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Magnetic Field
Magnetic Field The problem: A square conducting loop with sides L and carrying current I is lying in the x − y plane, so that its center is at the origin. Find the field on the z axis. The solution: The field of a finite wire at distance r is given by ~ = dB µ0 Id~l × (~r − ~r0 ) 4π |~r − ~r0 |3 (1) where ~r = (r, 0, 0), ~r0 = (0, 0, z 0 ) and d~l = (0, 0, dz 0 ) in the cylindrical coordinates (where we think of the wire going into the ẑ 0 direction.) The field is in ϕ̂ direction (around each wire) L µ0 I Bϕ = r 4π Z2 (r2 1 1 µ0 I L dz 0 = ϕ̂ 3/2 02 2 4π r (r + (L/2)2 )1/2 +z ) (2) −L 2 There are four wires that contribute to the field and we are interested in the ẑ component, so we have to take four times the field we found and to consider the angle between ẑ and field’s direction Bz = 4Bϕ cos θ = 4Bϕ L 2r (3) where r2 = ( L2 )2 + z 2 . For the projection onto the x − y plane see the Fig. below. (on the Fig. the angle between r and B is π/2 because B is a tangent). Finally Bz = µ0 2L2 I 1 q 4π ( L2 )2 + z 2 2( L )2 + z 2 (4) 2 1 Magnetic Field The problem: Two parallel half infinite wires are connected to a half ring wire with a radius R and carrying current I. The half loop and the wires are in the x − y plane. What is the magnetic field in the center of the half ring? The solution: This system can be duplicated and the duplicate system is put onto the original system, so that one has two infinite wire and a ring between them and tangent to them. The field B created in the center of the ring is twise the field created in the center of the half ring of the original system. The field in the center of a ring is B= µ0 I 2r (1) The field of infinite wire is B= µ0 I 2πr (2) Both wires contribute in the same direction (ẑ)as the loop, therefore using superposition and dividing by two, we get µ0 I µ0 I µ0 I 1 2 + = (2 + π) (3) B= 2 2πr 2r 4πr 1 Magnetic Field The problem: In a ring with radius R an arc of an angle α has been replaced with a straight cord. In the ring flows current I1 . A straight infinite wire passes through the center of the ring and carries a current I2 . Calculate the force and the torque on the ring. The solution: Using the superposition principle we calculate the forces due to the infinite wire acting on the cord and on the arc of an angle 2π − α and then sum them up. The force acting on the arc The magnetic field induced by the infinite wire is in the φ̂ direction and so is the element d~l = rdφφ̂. Then the force and the torque are zero. The force acting on the cord Using the Biot - Savart’s law we can calculate the field on the cord Z ~ µ0 dl × (~r − ~r0 ) ~ B = I 4π |~r − ~r0 |3 (1) Let the origin be at the center of the ring. The y-axis goes from the center of the ring toward the cord and perpendicular to it. So that we are interested in the field at α ~r = (x, R cos , 0) (2) 2 α α (3) −R sin < x < +R sin 2 2 The position of the current element d~l is ~r0 = (0, 0, z) d~l = dz ẑ (4) (5) Then α ~r − ~r0 = (x, R cos , −z) 2 r α ~r − ~r0 = x2 + R2 cos2 + z 2 2 α d~l × (~r − ~r0 ) = dz(−R cos x̂ + xŷ) 2 (6) (7) (8) Since the field in the x direction does not induce force, we calculate only the field in the y direction Z ∞ xdz 2KI2 x ~ By = KI2 (9) 3 = 2 x + R2 cos2 α2 −∞ (x2 + R2 cos2 α + z 2 ) 2 2 1 where K = µ0 4π . Now we need to calculate the force and the torque: ~ = Id~l × B d~τ = ~r × dF~ dF~ (10) (11) Then the force is dF~ F~ = −I1 dxx̂ × By ŷ = Z = dF~ = − R sin α 2 −R sin α 2 Z 2KI1 I2 x + R2 cos2 dxẑ (12) I1 dxBy (x)ẑ = 0 (13) x2 α 2 The integral is zero since the the integrand is an odd function. The torque (relatively to the middle of the cord) is 2KI1 I2 x2 dxŷ + R2 cos2 α2 Z Z R sin α 2 x2 dx ~τ = d~τ = 2I1 I2 K ŷ 2 2 2 −R sin α x + R cos d~τ = ~x × dF~ = (14) x2 2 2 α 2 α α α = 4I1 I2 KR sin − cos ŷ 2 2 2 (15)