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3 years ago

Compare how magnetic forces act through non-magnetic materials andmagnetic materials:

Physics

1 answer:

gavmur [86]3 years ago

5 0

<h2>Compare how magnetic forces act through non-magnetic materials and </h2><h2>magnetic materials:</h2>

Explanation:

Magnet

• Magnet :- is an object which attracts pieces of iron, steel etc towards itself.

Some facts about magnets:-

• When magnet is freely suspended it always align towards north-south direction

• Like poles always repel & opp. poles attract each other.

• Magnet always exist as dipole

• Two poles can never be separated : if we try to cut it then still both the poles will exist even ina small piece of magnet .it automatically develops the lost polarity

Magnet always develop certain area around it where its effect can be felt ie. magnetic field.

MAGNETIC Field

is studied by drawing imaginary lines called magnetic lines of forces.

Characteristics.

• They always originate from North pole & terminate at South pole. This shows that if north pole was free is move it would have mvre towards south pole.

•Place where they are closer indicate strong M. field i.e. at poles.

•Mag. Field lines gives the direction of magnetic force.

•Two magnetic lines will never intersect each other as they give direction of force & force can’t have 2 direction at a time.

M Field lines are closed continuous curves.

This is what that happens in magnetic materials .

Non magnetic materials

Magnetic forces 'act through' non-magnetic materials
These magnetic materials can be used as a shield around a magnet.
The domain theory of magnetism tries to explain why metals get magnetised
. The magnetic elements have little molecular magnets inside them.
Magnets attract only the magnetic materials

Types of magnetic materials

Soft magnetic materials (e.g. iron) have domains that easily move into line when the metal is placed in a magnetic field but as soon as the field is removed the domains take on a random pattern again. It returns to being unmagnetized straight away.

Hard magnetic materials (e.g. steel) have domains that do not easily move into line when the metal is placed in a magnetic field, a strong field is needed for some time, but then, when the field is removed the domains retain the magnetic pattern. The metal stays magnetic for a long time.

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ANTONII [103]

Answer:

Explanation:

a) ωp = 2π radians / 1.7 s = <u>3.7 rad/s</u>

b) ωs = 3.7 rad/s(9.5 cm / 4.5 cm) = 7.8 rad/s

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c) ωs = 3.5 m/s / 0.65 m = 5.38 rad/s

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3 years ago

An inverse-time circuit breaker (CB) is used for branch-circuit short-circuit and ground-fault protection for a 30-horsepower, 2

Maru [420]

Answer:

Explanation:

Motor rating is given in horsepower (hp), it will be converted in watt (W).

Standard to install circuit breaker for an electric circuit is usually 20% ~ 25% more than the Rated Current of the circuit

while

Standard to install overload relay for an electric circuit is usually 20% ~ 25% more than the Running Current of the circuit.

So, to find the maximum capacity of the circuit breaker, rated current of the motor will be multiplied by 1.2 ~ 1.25

Step by Step Explanation:

30hp = 22371W (as 1hp = 745.7)

Assuming unity power factor (cosФ=1) and 208V phase to phase voltage:

Rated Power (watt) = √3 . V.I. cosФ

<em>{if 208V is phase to neutral voltage, then use following formula:</em>

<em> Rated Power (watt) = 3 . V.I. cosФ}</em>

$\frac{22371}{\sqrt{3} * 208 * 1} = I$

Rated Current = <u>62.169A</u>

So, required maximum rating for circuit breaker is:

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3 years ago

A large sheet of charge has a uniform charge density of 9  μCm2. What is the electric field due to this charge at a point just

Alex73 [517]

Answer:

Explanation:

Surface charge density, σ = 9 μC/m² = 9 x 10^-6 C/m²

According to the Gauss theorem,

Electric field due to the sheet is given by

$E = \frac {\sigma }{2\epsilon _{0}}$

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3 years ago

A cylindrical resistor element on a circuit board dissipates 1.2 W of power. The resistor is 2 cm long, and has a diameter of 0.

34kurt

Answer:

(a) The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) The fraction of heat dissipated from the top and bottom surfaces is 0.045.

Explanation:

(a) The amount of heat dissipated ( $Q$ ), measured in joules, by the cylindrical resistor is the power multiplied by operation time ( $\Delta t$ ), measured in hours. That is:

$Q = \dot Q \cdot \Delta t$ (1)

If we know that $\dot Q = 1.2\,W$ and $\Delta t = 86400\,s$ , then the amount of heat dissipated by the resistor is:

$Q = (1.2\,W)\cdot (86400\,s)$

$Q = 103680\,J$

The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux ( $Q'$ ), measured in watts per square meter, is the heat transfer rate divided by the area of the cylinder ( $A$ ), measured in square meters:

$Q' = \frac{\dot Q}{A}$ (2)

$Q' = \frac{\dot Q}{\frac{\pi}{2}\cdot D^{2}+\pi\cdot D \cdot h }$ (3)

Where:

$D$ - Diameter, measured in meters.

$h$ - Length, measured in meters.

If we know that $\dot Q = 1.2\,W$ , $D = 4\times 10^{-3}\,m$ and $h = 2\times 10^{-2}\,m$ , the heat flux of the resistor is:

$Q' = \frac{1.2\,W}{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2}+\pi\cdot (4\times 10^{-3}\,m)\cdot (2\times 10^{-2}\,m) }$

$Q' \approx 4340.589\,\frac{W}{m^{2}}$

The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) Since heat is uniformly transfered, then the fraction of heat dissipated from the top and bottom surfaces ( $r$ ), no unit, is the ratio of the top and bottom surfaces to total surface: