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Delvig [45]
3 years ago
11

(a) A long, straight solenoid has N turns, uniform cross-sectional area A, and length l. Show that the inductance of this soleno

id is given by the equation L=μ0AN2/l. Assume that the magnetic field is uniform inside the solenoid and zero outside. (Your answer is approximate because B is actually smaller at the ends than at the center. For this reason, your answer is actually an upper limit on the inductance.) (b) A metallic laboratory spring is typically 5.00 cm long and 0.150 cm in diameter and has 50 coils. If you connect such a spring in an electric circuit, how much self-inductance must you include for it if you model it as an ideal solenoid?
Physics
1 answer:
Paul [167]3 years ago
6 0

Answer:

a. L = μ₀AN²/l b. 1.11 × 10⁻⁷ H

Explanation:

a. The magnetic flux through the solenoid, Ф = NAB where N = number of turns of solenoid, A = cross-sectional area of solenoid and B = magnetic field at center of solenoid = μ₀ni where μ₀ = permeability of free space, n = number of turns per unit length = N/l where l = length of solenoid and i = current in solenoid.

Also, Li = Ф where L = inductance of solenoid.

So, Li = NAB

= NA(μ₀ni)

= NA(μ₀Ni/l)

Li = μ₀AN²i/l

dividing both sides by i, we have

So, L = μ₀AN²/l

b. The self- inductance, L = μ₀AN²/l where

A = πd²/4 where d = diameter of solenoid = 0.150 cm = 1.5 × 10⁻³ m, N = 50 turns, μ₀ = 4π × 10⁻⁷ H/m and l = 5.00 cm = 5 × 10⁻² m

So, L = μ₀AN²/l

L = μ₀πd²N²/4l

L = 4π × 10⁻⁷ H/m × π(1.5 × 10⁻³ m)²(50)²/(4 × 5 × 10⁻² m)

L = 11,103.3 × 10⁻¹¹ H

L = 1.11033 × 10⁻⁷ H

L ≅ 1.11 × 10⁻⁷ H

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Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is  F_{net}=  3.22*10^{-5} \ J

Explanation:

From the question we are told that

    The third charge is  q_3 =  55 nC =  55 *10^{-9} C

    The position of the third charge is  x = -1.220 \ m

     The first charge is q_1 =  -16 nC  =  -16 *10^{-9} \ C

     The position of the first charge is x_1 =  -1.650m

      The second charge is  q_2 =  32 nC  =  32 *10^{-9} C

      The position of the second charge is  x_2 =   0  \ m  

The distance between the first and the third charge is

      d_{1-3} =  -1.650 -(-1.220)

     d_{1-3} = -0.43 \ m

The force exerted on the third charge by the first is  

     F_{1-3} =  \frac{k  q_1 q_3}{d_{1-3}^2}

Where k is the coulomb's constant with a value  9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.

substituting values

      F_{1-3} =  \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}

       F_{1-3} = 4.28 *10^{-5} \ N

 The distance between the second and the third charge is      

  d_{2-3} =  0- (-1.22)

   d_{2-3} =1.220 \ m

The force exerted on the third charge by the first is mathematically evaluated as

       F_{2-3} =  \frac{k  q_2 q_3}{d_{2-3}^2}

substituting values

       F_{2-3} =  \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}

       F_{2-3} =  1.06*10^{-5} N

The net force is

      F_{net} =  F_{1-3} -F_{2-3}

substituting values

    F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}

    F_{net}=  3.22*10^{-5} \ J

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