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dexar [7]
3 years ago
13

If 2.0 x 10^-4 C charge passes a point in 5.0 x 10^-5 s, what is the rate of current flow?

Physics
2 answers:
vampirchik [111]3 years ago
8 0
Current is defined as the rate of charge flowing a point every second. Having a current of 1 Ampere signifies 1 Coulomb is flowing in a circuit every second. It is measured by the use of an ammeter which is positioned in series to the component to be measured. The current in the problem is calculated as follows:

I = 2.0 x 10^-4 C / 5.0 x 10^-5 s
<span>I = 4 A</span>
VladimirAG [237]3 years ago
6 0

Q= I×t

Where Q is charge in Coulombs C,

I is current in Amperes A,

t is in seconds s.

The rate of current flow is defined as the charge passing through a point in the circuit per second.

Rearranging formula gives I=\frac{Q}{t}

⇒I = \frac{2.0*10^-4}{5.0*10^-5} = 4 Amperes

∴ current flow = 4.0×10^0 A

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In the first situation, the magnet is getting closer to the loop, so the magnetic flux through the area enclosed by the wire is increasing (because the intensity of the magnetic field B is increasing). Situation 2) is the opposite case: the wire loop is moving away from the magnet, so the intensity of the magnetic field B is decreasing, and therefore the magnetic flux is decreasing as well.
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Question 9
egoroff_w [7]

Answer:

C

Explanation:

F=ma

given solution

v=12m/s a=v/t

s=6 sec =12m/s÷6sec

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Two parallel wires separated by a distance of 0.6 m each carry current in the same direction. One wire is carrying a current of
mart [117]

Answer:

The value is  B =  3.33 *10^{-6} \  T

Explanation:

From the question we are told that

  The distance of separation is  d = 0.6 \  m

  The current on the one wire is I_1 =  9 \  A

  The current on the second wire is I_2 =  4 \ A

Generally the magnitude of the field exerted between the current carrying wire is

        B  =  B_1 - B_2

Here B_1 is the magnetic field due to the first wire which is mathematically represented as

         B_1 = \frac{\mu_o * I_1 }{2 \pi * d_1}

Here d_1 is the distance to the half way point of the separation and the value is  

    d_1 =  0.3 \  m

B_2 is the magnetic field due to the first wire which is mathematically represented as

         B_2  = \frac{\mu_o * I_2 }{2 \pi * d_2}

Here d_2 is the distance to the half way point of the separation and the value is  

    d_2 =  0.3 \  m  

This means that d_1 = d_2 = a =  0.3

So

     B =  \frac{\mu_o * I_1 }{2 \pi * d_1}  -  \frac{\mu_o * I_2 }{2 \pi * d_2}

=>  B =  \frac{\mu_o * (I_1 - I_2)}{2 \pi *0.3 }

=>  B =  \frac{  4\pi * 10^{-7}  * (9- 4)}{2 * 3.142  *0.3 }

=>  B =  3.33 *10^{-6} \  T

5 0
3 years ago
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