Please answer the \"Exercise\" portion questions. Please explain your answers le

Question

Please answer the "Exercise" portion questions. Please explain your answers legibly and step by step .

EXAMPLE 20.7 Inductance, Self-Induced emf, and Solenoids GOAL Calculate the inductance and self-induced emf of a solenoid PROBLEM (a) Calculate the inductance of a solenoid containing 300 turns if the length of the solenoid is 25.0 cm and its cross-sectional area is 4.00 x 104 m2. (b) Calculate the self-induced emf in the solenoid described in part (a) if the current in the solenoid decreases at the rate of 50.0 A/s STRATEGY Substituting given quantities into the equation (1) below gives the inductance L. For part (b), substitute the result of part (a) and ????--50.0 A/s into equation (2) to get the self-induced emf. (2) ?--L(Al/dt) SOLUTION (A) Calculate the inductance of the solenoid. Substitute the number V of turns, the area A, and the length into the equation to the right to find thee inductance (300 2(4.00 x 104 m2) 2.50 x 10-1 m HoN2A - 1.81 x 10-4 T-m2/A 0.181 mH

Explanation / Answer

Part A

Inductance of solenoid is given by:

L = u0*N^2*A/l

N = sqrt (L*l/(u0*A))

N = sqrt (0.282*10^-3*31.7*10^-2/(4*pi*10^-7*6.31*10^-4))

N = 335.76 turns = 336 turns

Now turns/meter will be

n = N/l = 335.76/0.317 = 1059 turns/m

Part B

We know that EMF is given by

EMF = -L*di/dt

di/dt = -EMF/L

Using given values:

di/dt = -(-13.8*10^-3)/(0.282*10^-3)

di/dt = 48.94 A/s

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