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Alisiya [41]
3 years ago
13

A long solenoid that has 1 200 turns uniformly distributed over a length of 0.420 m produces a magnetic field of magnitude 1.00

10-4 T at its center. What current is required in the windings for that to occur
Physics
1 answer:
Fantom [35]3 years ago
8 0

Answer:

<h2>The current required  winding is  2.65*10^-^2 mA</h2>

Explanation:

We can use the expression B=μ₀*n*I-------1 for the magnetic field that enters a coil  and

n= N/L (number of turns per unit length)

Given data

The number of turns n= 1200 turns

length L= 0.42 m

magnetic field B= 1*10^-4 T

μ₀= 4\pi*10^-^7 T.m/A

Applying the equation  B=μ₀*n*I

I= B/μ₀*n

I= B*L/μ₀*n

I= \frac{1*10^-^4*0.42}{4\pi*10^-^7*1.2*10^3 }

I= 2.65*10^-^2 mA

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The periodic table lists all the elements according to the number of __________ they have! You can’t change __________ without c
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Answer:

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2 years ago
If a car's engine gives off 65% thermal energy, what is the maximum efficiency?
stich3 [128]

Answer:

35%

Explanation:

The car's engine gives off 65% thermal energy

So only 35 % is converted into mechanical energy .

input heat = Q₁ = 100

output heat = Q₂ = 65

Work output = Q₁ - Q₂ = W

W = 100 - 65 = 35

Efficiency = W / Q₁ X 100

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A particle (mass = 2.0 mg, charge = −6.0 μC) moves in the positive direction along the x axis with a velocity of 3.0 km/s. It en
Virty [35]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The acceleration would be  a = 0.003* 6 =0.018\  m/s^2    

Explanation:

The objective of this solution is to obtain the acceleration of the particle

       Now looking at Newton law which is mathematically represented as

                       F = ma

  Where F is the force experience by a particle

              m is the mass of the particle

              a is the acceleration of the particle

  And also this force is equivalent to magnetic force in a magnetic field which is mathematically represented as

                    F = qvB

Where q is the charge of the particle

             v is the velocity of the  charge

             B is the magnetic field the charge is under it influence

  Now equating this two formulas

                   ma = qvB

 Making a the subject we have

                  a = \frac{qvB}{m}

In the question the direction of the is in the positive x-axis which is i hence the direction would be in the i direction

      So substituting  (2.0i +3.0j+4.0k)mT = (2.0i +3.0j+4.0k)*10^{-3}T for B

                    a = \frac{q}{m} * v (2.0i +3.0j +4.0k)*10^{-3}

      Substituting       3.0 Km /s = 3.0*10^{3}\ m/s  for v  and -6.0 \muC = -6.0*10^{-6} C for q

                     a = \frac{-6.0*10^{-6}}{2.0*10^{-3}} * 3.0*10^{3} *(2i+3j+4k) *10^{-3}

                       a = 0.003 * 3i(2i+3j+4k)

                      a = 0.003 *((3*2)i \ \cdot i \ +(3*3) i \ \cdot \ j  \ + (3*4)i \ \cdot \ k)

According to vector multiplication

                                             i \cdot i = j \cdot j = k\cdot k = 1\\\\and \ i\cdot j = i\cdot k  = 0

     So

               a = 0.003* 6 =0.018\  m/s^2          

     

8 0
3 years ago
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