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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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Answer:

Explanation:

First, It's important to remember F = ma, and in this problem m = 13.3 kg

This can be reduced to a simple system of equations problem.  Now if they are both going the same way then we add them, while if they are going the opposite way we subtract them.  So let's call them F1 and F2, with F1 arger than F2.  Now, When we add them together F1+F2 = (.723 m/s^2)*13.3kg and then when we subtract them, and have the larger one pushing toward the east, let's call F1 the larger one, F1-F2 = (.493 m/s^2)*13.3kg.  

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