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Mademuasel [1]
3 years ago
6

A wire along the z axis carries a current of 4.9 A in the z direction Find the magnitude and direction of the force exerted on a

3.3 cm long length of this wire by a uniform magnetic field pointing in the x direction having a magnitude 0.43T
Physics
1 answer:
AURORKA [14]3 years ago
5 0

Answer:

<h2>0.069 N, in the X direction</h2>

Explanation:

According to Flemming's left hand rule, it sates that if the first three fingers of the left hand are held mutually at right angles to one another, the fore finger will point in the direction of magnetic field, the middle finger will point in direction of current, while the thumb will point to the direction of force.

Mathematically the law is stated as

F= BIL

given data

Magnetic field B=  0.43T

Current I= 4.9 A

length of conductor L= 3.3cm to meter , 3.3/100=  0.033 m

Applying the formula the force is calculated as

F= 0.43*4.9* 0.033= 0.069 N

According to Flemming's rule the direction of all parameters are mutually perpendicular to one another, then the Force is in the X direction

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What is a difference in electrical charge from one point to another called?
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Answer:electrical potential

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When two objects collide their momentum after the collision is explained by what?
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<span>When two objects collide their momentum after the collision is explained by</span> the conservation of momentum
5 0
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A charged particle accelerated to a velocity v enters the chamber of a mass spectrometer. The particle's velocity is perpendicul
gladu [14]

Answer:

Circle

Explanation:

When a charged particle is in motion in a region with magnetic field, the particle experiences a force whose magnitude is given by

F=qvB sin \theta

where

q is the charge

v is the velocity of the particle

B is the strength of the magnetic field

\theta is the angle between the directions of v and B

In this problem, the velocity of the particle is perpendicular to the magnetic field, so

\theta=90^{\circ}

and the formula reduces to

F=qvB

Also, the direction of this force is perpendicular to the direction of motion of the particle. This means that as the charge moves in the region of the magnetic field, the force acting on it acts as a centripetal force: therefore, the particle will start moving by unifom circular motion, with constant speed (because the magnetic force does no work on the particle, since it is perpendicular to the direction of motion).

So, the path of the particle will be a circle.

4 0
3 years ago
A total charge of 9.0 mC passes through a cross-sectional area of a nichrome wire in 3.6s. The number of electrons passing throu
Setler79 [48]
<h2>Given :</h2>

  • total charge = 9.0 mC = 0.009 C

Each electron has a charge of :

1.6 \times 10 {}^{ - 19} \:  C

For producing 1 Cuolomb charge we need :

  • \mathrm{\dfrac{1}{1.6 \times 10 {}^{ - 19} } }

  • \dfrac{10 {}^{19} }{1.6}

  • \dfrac{10\times 10 {}^{19} }{16}

  • \dfrac{100 \times 10 {}^{18} }{16}

  • \mathrm{6.24 \times 10 {}^{18}  \:  \: electrons}

Now, for producing 0.009 C of charge, the number of electrons required is :

  • 0.009 \times 6.24 \times  {10}^{18}

  • 0.05616 \times 10 {}^{18}

  • \mathrm{5.616 \times 10 {}^{16}  \:  \: electons}

_____________________________

So, Number of electrons passing through the cross section in 3.6 seconds is :

\mathrm{5.616 \times 10 {}^{16} \:  \: electrons}

Number of electrons passing through it in 1 Second is :

  • \dfrac{5.616 \times  {10}^{16} }{3.6}

  • \mathrm{1.56 \times 10 {}^{16}  \:  \: electrons}

Now, in 10 seconds the number of electrons passing through it is :

  • 10 \times  \mathrm{1.56 \times 10 {}^{16}  \:  \: }

  • \mathrm{1.56 \times 10 {}^{17}  \:  \: electrons}

_____________________________

\mathrm{ \#TeeNForeveR}

6 0
3 years ago
A pursuit spacecraft from the planet Tatooine is attempting to catch up with a Trade Federation cruiser. As measured by an obser
Ede4ka [16]

Answer:

0.384c

Explanation:

To find the speed of the pursuit ship relative to the cruiser you use the following relativistic equation:

u'=\frac{u-v}{1-\frac{uv}{c^2}}

u': relative speed

u: speed of the pursuit ship = 0.8c

v: speed of the cruiser = 0.6c

c: speed of light

You replace the values of the parameters to obtain u':

u'=\frac{0.8c-0.6c}{1-\frac{(0.6c)(0.8c)}{c^2}}=0.384c

Hence, the relative speed is 0.384c

4 0
3 years ago
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