Question

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Two long parallel wires, P and Q are fixed on a horizontal plane north-south direction. Wire P is 10 cm left of wire Q. Wire P carries 40 A current northwards (a) while wire Q carries 20 A current southwards. Determine the magnitude and direction of i. the resultant magnetic field at midpoint between the wires. the resultant magnetic field at a point on the horizontal plane, 8.0 cm from the wire P and 18 cm from wire Q the magnitude of the force per unit length on each wire. 6 marks] A 2.0 m long wire is made into a plane coil of five turns. If the current in the wire is 30 mA, ) i. calculate the magnetic field at the centre of the coil. ii. show the direction of the magnetic field by a simple sketch. [4 marks

Answer 1

(a) (i) Magnetic field at midpoint between the wires P & Q B_1 = B + p + B_Q = mu_0/2 pi I_P/r_P + mu_0/2 pi I_Q/r_Q B_1 = 2 times 10^-7 [40/0.05 + 20/0.05] T B_1 = 2.4 times 10^-4 T or, B = 0.24 mT (ii) Magnetic field at point (2) B_2^Vector = B_P - B_Q = mu_0/2 pi [i_P/r_P - I_Q/r_Q] B_2^Vector = 2 times 10^-7 [40/0.08 - 20/0.18] T B_2^Vector = 7.78 times 10^-5 T or, B_2 = 0.0778 mT (iii) Magnitude of force per unit length, dF/dL = mu_0/2 pi I_P I_Q/r Therefore dF/dL = 2 times 10^-7 times 40 times 20/0.1 N/m dF/dL = 1.6 times 10^-3 N/m (b) (i) Radius of coil is given by M(2 pi r) = L = > 5 (2 pi r) = 2 r = 1/5 pi m Therefore Magnetic field at centre, B = n mu_0 I/2r Therefore B = 5 times 4 pi times 10^-7 times 30 times 10^-3