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3 years ago

Which is true of magnetic field lines but not electric field lines?

2 answers:

7 0

The closer the field lines, the stronger the field, whereas for electric field, the higher the electric flux passing through, the stronger the electric field.

Explanation:

Another difference between the two is that magnetic fields formed closed loops around the magnetic while electric fields do not. Electric fields are measured in two dimensional while magnetic fields in 3 dimensional. The electric field line can do work while magnetic fields can not (because particles in magnetic field remain constant even though the charge may change direction).

Learn More:

For more on electric and magnetic fields check out;

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11Alexandr11 [23.1K]3 years ago

7 0

Answer:

A.) They show which way iron shavings would align themselves.

Explanation:

APEX:)))

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Two charged concentric spherical shells have radii of 11.0 cm and 14.0 cm. The charge on the inner shell is 3.50 ✕ 10−8 C and th

Sergio039 [100]

Answer:

The magnitude of the electric field are $2.38\times10^{4}\ N/C$ and $1.09\times10^{4}\ N/C$

Explanation:

Given that,

Radius of inner shell = 11.0 cm

Radius of outer shell = 14.0 cm

Charge on inner shell $q_{inn}=3.50\times10^{-8}\ C$

Charge on outer shell $q_{out}=1.60\times10^{-8}\ C$

Suppose, at r = 11.5 cm and at r = 20.5 cm

We need to calculate the magnitude of the electric field at r = 11.5 cm

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Where, q = charge

k = constant

r = distance

Put the value into the formula

$E=\dfrac{9\times10^{9}\times3.50\times10^{-8}}{(11.5\times10^{-2})^2}$

$E=2.38\times10^{4}\ N/C$

The total charge enclosed by a radial distance 20.5 cm

The total charge is

$q=q_{inn}+q_{out}$

Put the value into the formula

$q=3.50\times10^{-8}+1.60\times10^{-8}$

$q=5.1\times10^{-8}\ C$

We need to calculate the magnitude of the electric field at r = 20.5 cm

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times5.1\times10^{-8}}{(20.5\times10^{-2})^2}$

$E=1.09\times10^{4}\ N/C$

Hence, The magnitude of the electric field are $2.38\times10^{4}\ N/C$ and $1.09\times10^{4}\ N/C$

7 0

3 years ago

The era of planet formation ended when the remaining hydrogen and helium gas of the solar nebula was swept into interstellar spa

artcher [175]

Answer:

The era of planet formation ended when the remaining hydrogen and helium gas of the solar nebula was swept into interstellar space by the solar winds.

Explanation:

The Solar System is formed from a molecular cloud (compound by gas and dust). If there is a near perturbation to the cloud, maybe due to a supernova explosion, the molecular cloud will collapse under its own gravity. Then, in some point it starts to rotate and will accrete all the material in a disk around the protostar¹.

Inside the disk, dust particles start to collide and accrete until they form planetesimals². As a consequence of the gravitational force of the star, rocky and metallic particles will be more attracted to the inner part of the Solar System (close to the Sun) since they have more mass than gas.

Then, when the star has the necessary pressure and temperature to initiate nuclear reactions in its core, it will be able to emit huge amounts of energy, better known as solar winds. These winds will expel gas (hydrogen and helium) from the Solar System more easily than the rocky and metallic particles.

Notice that when such event occurs, rocky and gaseous planets were already formed.

Key terms:

¹Protostar: A young star.

²Planetesimals: Object formed by many fragments due to the gravitational attraction between them.

5 0

3 years ago

A plane traveled west for 4.0 hours and covered a distance of 4,400 kilometers. What was its velocity? 18,000 km/hr

olga55 [171]

Velocity is the same as the formula for speed, the only difference is velocity has direction. Velocity is distance over time. Given is 4,400 kilometers travelled west in 4 hours. Applying the given equation, we wil have 4,400/4 = 1,100 km/hr west

5 0

3 years ago

Read 2 more answers

The area vector of a square loop of 5 turns of a conductor each with side length of 0.2 m carrying a current of 2 A is antiparal

Anon25 [30]

Answer:

$M= 0.4 Am^2$