A 20-turn wire coil (i,e. like the solenoid station lab) having radius 2.5 cm ha

Question

A 20-turn wire coil (i,e. like the solenoid station lab) having radius 2.5 cm has total resistance 5.0 The coil cm from the wire, enough away that you may consider all part of the coil to be 50 cm from the wire. The current in the wire is steadily increased 0 A to 10 A in 30 sec. When the current in the wire has reached 10 A. what a the and direction. of the wire's magnetic field at a point inside the 20 loops? While the wire current is increasing, compute the magnitude and of the current induced in the coil

Explanation / Answer

Given that Number turns N = 20 radius r =2.50 x 10^-2 m resistance R = 5.0 I/t = (10 A- 0)/(30.0 s) d = 0.50 m ------------------------------------------------------------ The magnetic field due to long straight wire is           B = oI/(2d) The magnitude of induced emf is             e = NA (B/t)                = NA (o/(2d))(I/t)                = (20)((2.50 x 10^-2 m)^2)[(4 x10^-7 H/m)/(2(0.50 m))]((10 A- 0)/(30.0 s))                = 5.24x10^-9 V The magnitude of induced current is           I = e/R              = ( 5.24x10^-9 V)/(5.0 )             = 1.05 x10^-9 A Direction : clockwise. Direction : clockwise.                

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