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lbvjy [14]
3 years ago
10

Suppose that the current in the solenoid is i(t. within the solenoid, but far from its ends, what is the magnetic field b(t due

to this current?

Physics
2 answers:
Serjik [45]3 years ago
8 0

The magnetic field b(t) due to this current is B(t)=mu_0*n*I(t)

<h3>Explanation: </h3>

Suppose that the current in the solenoid is i(t). Within the solenoid, but far from its ends, what is the magnetic field B(t) due to this current?

The magnetic field is the area around a magnet where there is magnetic force.  Moving electric charges can make magnetic fields. Then a solenoid is the long coil of wire wrapped in many turns. When a current passes through, it creates a nearly uniform magnetic field inside.  

Solenoids can convert electric current to mechanical action, and so are very commonly used as switches

Even small solenoids can exert forces of a few newtons.

If we look through the solenoid far from ends, we can use Ampere's law to calculate the field strength. The magnetic field (deep) within the solenoid has a uniform value B, and outside the coils has value zero.

The magnetic field within a solenoid depends upon the current and density of turns. The magnetic field B(t) due to this current is B(t)=mu_0*n*I(t). This is derived from Ampere's law used to calculate the strength of magnetic field.

Where n is number of coils per meter and I is current through wire.

Learn more about the magnetic field  

brainly.com/question/12450147

#LearnWithBrainly

Mkey [24]3 years ago
7 0
The answer is B(t) = constants x I(t)

Please take precaution on the point that it is an independent field of its radial position, if the point is measured well in the solenoid. (also the radial position is the axis of its symmetry)
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qwelly [4]

Answer:

59.503987 seconds

Explanation:

b = Proportionality constant = 50 Ns/m

g = Acceleration due to gravity = 9.81 m/s²

m = Mass of object = 700 kg

We have the equation of velocity

v(t)=\dfrac{mg}{b}+\left(V_0-\dfrac{mg}{b}\right)e^{\dfrac{bt}{m}}

The equation of motion

x(t)=\dfrac{mg}{b}+\dfrac{m}{b}\left(V_0-\dfrac{mg}{b}\right)(1-e^{\dfrac{bt}{m}})

x(t)=\dfrac{700\times 9.81}{50}+\dfrac{9.81}{50}\left(0-\dfrac{700\times 9.81}{50}\right)(1-e^{\dfrac{50t}{700}})

when x(t)=2000

2000=\dfrac{700\times 9.81}{50}+\dfrac{9.81}{50}\left(0-\dfrac{700\times 9.81}{50}\right)(1-e^{\dfrac{50t}{700}})\\\Rightarrow 2000\times \:50=\frac{700\times \:9.81}{50}\times \:50+\frac{9.81}{50}\left(0-\frac{700\times \:9.81}{50}\right)\left(1-e^{\frac{50t}{700}}\right)\times \:50\\\Rightarrow 6867-\frac{67365.27}{50}\left(1-e^{\frac{50t}{700}}\right)=100000\\\Rightarrow 50\left(-\frac{67365.27}{50}\left(1-e^{\frac{50t}{700}}\right)\right)=93133\times \:50\\\Rightarrow \frac{-67365.27\left(1-e^{\frac{50t}{700}}\right)}{-67365.27}=\frac{4656650}{-67365.27}\\\Rightarrow 1-e^{\frac{50t}{700}}=-69.12538\dots\\\Rightarrow -e^{\frac{50t}{700}}=-70.12538\dots\\\Rightarrow t=14\ln \left(70.12538\dots \right)\\\Rightarrow t=59.50398\ s

The time taken is 59.503987 seconds

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levacccp [35]

Answer:

The coefficient of friction in the hall is 0.038

Explanation:

Given;

mass of the Parker, m = 73.2 kg

applied force on the parker, F = 123 N

frictional force, Fs = 27.4 N

the coefficient of friction in the hall = ?

frictional force is given by;

Fs = μN

Where;

μ is the coefficient of friction

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Where N is the Newton unit. N=kg ⋅
⋅
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2


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