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4 years ago

Using diagram differentiate between solenoid and a toroid

1 answer:

3 0

The Toroid is form when you have wound conductor around circular body. In this case you have magnatic field inside the core but you dont have any poles because circular body dont have ends. This can be used where you want minimum flux leakage and dont need magnatic poles. i.e. toroidal inductor, toroidal transformer.

The Solenoid is forn when you wound conductor around body with limb. In this case magnatic field creates two poles N and S. Solenoids have little bit flux leakage. This used where you want magnatic poles and flux leakage is not an issue. i.e. relay, motors, electromagnates.

1 == toroid

2= solenoid

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A horizontal rod 0.250 m long is mounted on a balance and carries a current. At the location of the rod a uniform horizontal mag

Karo-lina-s [1.5K]

Answer:

The current is 7.87 A.

Explanation:

Given that,

Length = 0.250 m

Magnetic field $B= 6.10\times10^{-2}\ T$

Magnetic force = 0.120 N

We need to calculate the current

Using formula of magnetic force

$F=BIL$

$I=\dfrac{F}{BL}$

Where, B = magnetic field

I = current

l = length

F = force

Put the value into the formula

$I=\dfrac{0.120}{6.10\times10^{-2}\times0.250}$

$I=7.87\ A$

Hence, The current is 7.87 A.

7 0

4 years ago

Which of the following factors might be affected by a wildfire? a. food b. shelter c. biodiversity d. all of the above Please se

maxonik [38]

Answer: d. all of the above

Explanation:

A wildfire can be either naturally occurring or induced by humans for their own benefit. Is likely to exert a negative impact on the wildlife. A fire is likely to spread quickly to the entire canopy cover of the forest trees. This will destroy the availability of food and other resources. Many birds and animals will loose their habitat or shelter. The destruction of the floral species will cause migration of the faunal species to other locations. Therefore, the biodiversity of the region will be lossed.

4 0

3 years ago

Read 2 more answers

Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. If the surface charg

bekas [8.4K]

Answer:

$5.3\times 10^3 N/C$

Explanation:

We are given that

Distance between plates=d=2.2 cm= $2.2\times 10^{-2} m$

$1 cm=10^{-2} m$

$\sigma=47nC/m^2=47\times 10^{-9}C/m^2$

Using $1 nC=10^{-9} C$

We have to find the magnitude of E in the region between the plates.

We know that the electric field for parallel plates

$E=\frac{\sigma}{2\epsilon_0}$

$E_1=\frac{\sigma}{2\epsilon_0}$

$E_2=\frac{\sigma}{2\epsilon_0}$

$E=E_1+E_2$

$E=\frac{\sigma}{2\epsilon_0}+\frac{\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}$

Where $\epsilon_0=8.85\times 10^{-12}C^2/Nm^2$

Substitute the values

$E=\frac{47\times 10^{-9}}{8.85\times 10^{-12}}$

$E=5.3\times 10^3 N/C$

Hence, the magnitude of E in the region between the plates= $5.3\times 10^3 N/C$

5 0

3 years ago

What is the acceleration of an object that slows from 15.2 ms to 11.2 ms in 3 seconds

Lelu [443]

Answer:

-1.333m/s2

Explanation:

a= v-u/t= 1.333m/s2

7 0

3 years ago

An object executing simple harmonic motion has a maximum speed of 4.3 m/s and a maximum acceleration of 0.65 m/s2 . find (a) the

Alexxx [7]

(1) The position around equilibrium of an object in simple harmonic motion is described by
$x(t) = A \cos (\omega t)$
where
A is the amplitude of the motion
$\omega$ is the angular frequency.

The velocity is the derivative of the position:
$v(t)=-\omega A \sin(\omega t) = -v_0 \sin (\omega t)$
where
$v_0 = \omega A$ is the maximum velocity of the object.

The acceleration is the derivative of the velocity:
$a(t)=- \omega^2 A \cos (\omega t) = -a_0 \cos (\omega t)$
where
$a_0=\omega^2 A$ is the maximum acceleration of the object.

We know from the problem both maximum velocity and maximum acceleration:
$v_0 = \omega A = 4.3 m/s$
$a_0 = \omega^2 A = 0.65 m/s^2$
From the first equation, we get
$A= \frac{4.3 }{\omega}$ (1)
and if we substitute this into the second equation, we find the angular fequency
$\omega=0.15 rad/s$
while the amplitude is (using (1)):
$A=28.7 m$

(b) We found in the previous step that the angular frequency of the motion is
$\omega=0.15 rad/s$
But the angular frequency is related to the period by
$T= \frac{2 \pi}{\omega}$
and so, the period is
$T= \frac{2 \pi}{\omega}= \frac{2 \pi}{0.15 rad/s}=41.9 s$

5 0

4 years ago