Chapter 22 Problems and Answers

# 9 x 10 5 telsapoints north 72 degrees below the

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magnetic field is 5.9 x 10^-5 Telsapoints North 72 degrees below the horizontal. Find the directionand magnitude of the magnetic force exerted on a 250m length ofwire if the current flows (a) horrizontally towards the east b) Horizontally towards the south a) Magnetic force acting on the wire when the current flowshorrizontally towards the east is F = BILsin θ = (5.9*10 -5 T)(110A)(250m)sin90 o Direction is points towards North, 18 o above thehorizontal. b) Magnetic force acting on the wire when the current flowshorizontally towards the south is F= BILsin θ = (5.9*10 -5 T)(110A)(250m)sin72 o Direction is points towards East. 22.43 Two currents loops, one square and the other circular have one turnmade from wires of the same length a)if these loops carry the same current and are placed in amagnetic field of equal

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magntidude is the maximum torque of thesquare loop greater than, less than or the same as the maximumtorque of the circular loop. Explain b) Calculate the ratio of the maximum torques τ square/ τ circle We know that maximum torque τ = NIAB N = 1 turn B = magnetic field I= current i n the loops let length of the wire be l Then perimeter of the square is 4a = l Then a = l/4 Area of the square is A = a 2 =(l/4) 2 circumference of the circle is 2 π r = l Then r = l/2 π area of the circlular loop A = π r 2 = π ( l/2 π ) 2 = l 2 /4 π Then τ square = 1*I*(l/4) 2 *B =BIl 2 /16 Now τ circle =1*I*(l 2 /4 π )*B = BIl 2 /4 π thus maximum torque on the square coil is less than maximumtorque on the circular coil.

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