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babunello [35]
3 years ago
14

A magnet in the form of a cylindrical rod has a length of 7.30 cm and a diameter of 1.5 cm. It has a uniform magnetization of 5.

30 x 103 A/m. What is its magnetic dipole moment in units of JIT?
Physics
2 answers:
yulyashka [42]3 years ago
6 0

Answer:

Magnetic dipole moment is 0.0683 J/T.

Explanation:

It is given that,

Length of the rod, l = 7.3 cm = 0.073 m

Diameter of the cylinder, d = 1.5 cm = 0.015 m

Magnetization, M=5.3\times 10^3\ A/m

The dipole moment per unit volume is called the magnetization of a magnet. Mathematically, it is given by :

M=\dfrac{\mu}{V}

\mu=M\times \pi r^2\times l

Where

r is the radius of rod, r = 0.0075 m

\mu=5.3\times 10^3\ A/m\times \pi (0.0075)^2\times 0.073\ m

\mu=0.0683\ J/T

So, its magnetic dipole moment is 0.0683 J/T. Hence, this is the required solution.

8_murik_8 [283]3 years ago
3 0

Answer:

0.068317J/T

Explanation:

Magnetization is defined as the magnetic moment per unit volume means

Magnetization=\frac{Magnetic\ moment}{Volume}

So magnetic moment = magnetization × volume

We have given magnetization = 5.3\times 10^3A/m

Diameter=1.5 cm ,so radius =\frac{1.5}{2}=0.75cm\ =0.0075m

Area A =\pi r^2=3.14\times 0.0075^2=1.766\times 10^{-4}m^2

Length = 7.3 cm =0.073 m

So volume =1.766\times10^{-4}\times 0.073=1.289\times 10^{-5}m^3

Now magnetic moment = magnetization × volume =5.30\times 10^3\times 1.289\times 10^{-5}=0.068317J/T

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Heat flows into a gas in a piston and work is performed on the gas by its surroundings. The amount of work done is equal to the
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Explanation:

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why does diving 30m below sea level affect our bodies more than being in a building 30m above sea level
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An alpha particle has a charge of +2e and a mass of 6.64 x 10-27 kg. It is accelerated from rest through a potential difference
kondor19780726 [428]

Answer:

a) v = 1.075*10^7 m/s

b) FB = 7.57*10^-12 N

c) r = 10.1 cm

Explanation:

(a) To find the speed of the alpha particle you use the following formula for the kinetic energy:

K=qV          (1)

q: charge of the particle = 2e = 2(1.6*10^-19 C) = 3.2*10^-19 C

V: potential difference = 1.2*10^6 V

You replace the values of the parameters in the equation (1):

K=(3.2*10^{-19}C)(1.2*10^6V)=3.84*10^{-13}J

The kinetic energy of the particle is also:

K=\frac{1}{2}mv^2       (2)

m: mass of the particle = 6.64*10^⁻27 kg

You solve the last equation for v:

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.84*10^{-13}J)}{6.64*10^{-27}kg}}\\\\v=1.075*10^7\frac{m}{s}

the sped of the alpha particle is 1.075*10^6 m/s

b) The magnetic force on the particle is given by:

|F_B|=qvBsin(\theta)

B: magnitude of the magnetic field = 2.2 T

The direction of the motion of the particle is perpendicular to the direction of the magnetic field. Then sinθ = 1

|F_B|=(3.2*10^{-19}C)(1.075*10^6m/s)(2.2T)=7.57*10^{-12}N

the force exerted by the magnetic field on the particle is 7.57*10^-12 N

c) The particle describes a circumference with a radius given by:

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the radius of the trajectory of the electron is 10.1 cm

6 0
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How much pressure is applied to the ground
statuscvo [17]

Answer:

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Explanation:

Pressure is defined as the perpendicular force applied per unit area.

i.e.  Pressure=\frac{Force}{Area}

Now, Force= mg

where, m = mass of the body(man) = 93 kg

g = acceleration due to gravity of Earth = 9.81 m/{s^{2}}

Area covered is equal to the area of both stilts(a man generally stands on two feet)

therefore Area=2(0.04)^{2} m^{2}

and putting in the values, we get,

Pressure=\frac{93\times9.81}{2\times(0.04)^{2}}Nm^{-2}=285103.125Nm^{-2}

Now we need to convert to our required units:

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Using the above relations we get,

Pressure=285103.125Pa=0.000145038\times285103.125lb/in^{2}=41.35lb/in^{2}

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