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3 years ago

the diagram to the right shows an electrical wire. the arrow shows the direction of the current. what does “b” represent?

Physics

1 answer:

goblinko [34]3 years ago

8 0

B represents the direction of the magnetic field around the wire

Explanation:

A wire carrying an electric current always produces a magnetic field around itself. The lines of the magnetic field produced by a current-carrying wires are concentric circles around the wire. The magnitude of the field is given by the formula:

$B=\frac{\mu_0 I}{2 \pi r}$

where

$\mu_0$ is the vacuum permeability

I is the current in the wire

r is the distance from the wire

The direction of the field lines is given by the so-called right hand rule, shown in the figure. Basically, the thumb of the right hand is placed in the direction of the electric current, while the other fingers are "wrapped" around the thumb: the direction of the other fingers give the direction of the magnetic field lines.

Learn more about magnetic field:

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A SDOF undamped system is set into free vibration with an initial displacement and an initial velocity. The mass of the system i

loris [4]

Answer:

Explanation:

stiffness k = 160

m = 10

angular frequency ω = $\sqrt{\frac{k}{m} }$

= $\sqrt{\frac{160}{10} }$

= 4

ω = 4

Let x = 4 - A sinωt

when t = 0

x = 4 in

when t = 2 s , x = - 4

- 4 = 4 - A sinωt

8 = A sin 4 x 2

8 = A sin8

A = 8 / sin 8

= 8 / .989

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x = 4 - A sinωt

dx / dt = - Aω cosωt

v = - Aω cosωt

for t = 0

v = - Aω

= - 8.09 x 4

= - 32.36 in / s

initial velocity v = - 32.36 in /s

displacement x for t = 4s

x = 4 - 8.09 sin 4 x 4

= 4 - 8.09 sin 16

= 4 - 8.09 x - .2879

= 4 + 2.33

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4 0

3 years ago

A solenoidal coil with 22 turns of wire is wound tightly around another coil with 330 turns. The inner solenoid is 21.0 cm long

Airida [17]

Answer:

$0.00027646\ T$

$2.33\times 10^{-5}\ H$

-0.04194 V

Explanation:

$N_2$ = Number of turns in outer solenoid = 330

$N_1$ = Number of turns in inner solenoid = 22

$I_1$ = Current in inner solenoid = 0.14 A

$\dfrac{dI_2}{dt}$ = Rate of change of current = 1800 A/s

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

r = Radius = 0.0115 m

Magnetic field is given by

$B=\mu_0\dfrac{N_2}{l}I\\\Rightarrow B=4\pi \times 10^{-7}\times \dfrac{330}{0.21}\times 0.14\\\Rightarrow B=0.00027646\ T$

The average magnetic flux through each turn of the inner solenoid is $0.00027646\ T$

Magnetic flux is given by

$\phi=BA\\\Rightarrow \phi=0.00027646\times \pi 0.0115^2\\\Rightarrow \phi=1.14862\times 10^{-7}\ wb$

Mutual inductance is given by

$M=\dfrac{N_1\phi}{I}\\\Rightarrow M=\dfrac{22\times 1.14862\times 10^{-7}}{0.14}\\\Rightarrow M=2.33\times 10^{-5}\ H$

The mutual inductance of the two solenoids is $2.33\times 10^{-5}\ H$

Induced emf is given by

$\epsilon=-M\dfrac{dI_2}{dt}\\\Rightarrow \epsilon=-2.33\times 10^{-5}\times 1800\\\Rightarrow \epsilon=-0.04194\ V$

The emf induced in the outer solenoid by the changing current inthe inner solenoid is -0.04194 V

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