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igomit [66]
3 years ago
5

1.) a current-carrying conductor of length 0.8 m is placed perpendicular to the direction of the magnetic field of 0.6 tesla. wh

at is the value of the current in the conductor if the conductor experiences a force of 1.2 n? 2.5 a 1.6 a 1.2 a 0.4 a 2.) an electron moves to the east and enters the magnetic field. the direction of the magnetic field is perpendicular to the direction of velocity acting downward. what is the direction of the force the electron experiences? east west north south
Physics
2 answers:
ella [17]3 years ago
7 0

The correct answer is:     south

BARSIC [14]3 years ago
4 0
1. The current can be calculated using the equation :

Force = current × length of wire × magnetic field strength

Plug in all the values

1.20 = current × 0.80 × 0.60

Simplify

1.20 = current × 0.48

Isolate for current

current = 2.5 Amps

2. electron moves east
magnetic field is downwards

the velocity, field and force all all perpendicular to each other. You can use the straight fingered left hand rule to determine the direction of the force.

Point your index finger to the right side (east), point your middle finger down towards the ground while you keep your index finger in place. You're left with your thumb pointing towards you, which is the south direction.

The direction of the force is south
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What is the largest four subsystems?​
AnnZ [28]

Explanation:

Everything in Earth's system can be placed into one of four major subsystems: land, water, living things, or air. These four subsystems are called "spheres." Specifically, they are the "lithosphere" (land), "hydrosphere" (water), "biosphere" (living things), and "atmosphere" (air).

5 0
3 years ago
An electric dipole is formed from ±1.00nC charges spaced 3.00 mm apart. The dipole is at the origin, oriented along the x-axis.
Ronch [10]

Answer:

Value of electric field along the axis and equitorial axis  E=31.25\ N/c and E = 15.625\ N/c respectively.

Explanation:

Given :

Distance between charges , d = 3 \ mm =\dfrac{3}{1000}\ m=3\times 10^{-3}\ m.

Magnitude of charges , q=1\ nC = 10^{-9}\ C.

Dipole moment , p=qL=10^{-9}\times 3\times 10^{-3}=3\times 10^{-12} \ C\ m.

Case A) (x,y) = (12.0 cm, 0 cm) :

Electric field of dipole in its axis ,

E=\dfrac{2kp}{r^3}

Putting all values and r=12\times 10^{-2}\ m.

We get , E=31.25\ N/c.

Case B) (x,y) = (0 cm, 12.0 cm) :

Electric field of dipole on equitorial axis ,

E = \dfrac{kp}{r^3}

Putting all values and r=12\times 10^{-2}\ m.

We get , E = 15.625\ N/c.

Hence , this is the required solution.

7 0
3 years ago
The surface charge density on an infinite charged plane is - 2.10 ×10−6C/m2. A proton is shot straight away from the plane at 2.
inn [45]

Explanation:

Formula to calculate electric field because of the plate is as follows.

         E = \frac{\sigma}{2 \times \epsilon_{o}}

            = \frac{2.10 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}}

           = 1.18 \times 10^{5} N/C

Now, we will consider that equilibrium of forces are present there. So,

                   ma = qE

       a = \frac{1.6 \times 10^{-19} \times 1.18 \times 10^{5}}{1.67 \times 10^{-27}}

          = 1.13 \times 10^{13} m/s^2

According to the third equation of motion,

         v^{2} = 2 \times a \times d

or,      d = \frac{v^{2}}{2d}

             = \frac{(2.4 \times 10^{6})^{2}}{2 \times 1.13 \times 10^{13}}

             = 0.254 m

Thus, we can conclude that the proton will travel 0.254 m before reaching its turning point.

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