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gavmur [86]
3 years ago
14

An object carries a charge of -6.1 µC, while another carries a charge of -2.0 µC. How many electrons must be transferred from th

e first to the second object so that both objects have the same charge? Number of electrons
Physics
1 answer:
Aleksandr [31]3 years ago
8 0

Answer: 2.05µC

Explanation:

First object carries a charge of -6.1µC Second object carries a charge of -2.0 µC.

To make the charges same, divide the difference between both charges by 2

i.e - 6.1µC - (-2.0µC)

= -6.1µC + 2.0µC

= -4.1µC

Then, divide -4.1µC by 2

i.e (-4.1µC) /2

= -2.05µC

Now, transfer -2.05µC from first object to second object

i.e (-6.1µC - (-2.05µC))

= - 6.1µC + 2.05µC

= - 4.05µC

Thus, the second object becomes

(-2.0µC + (-2.05µC))

= -2.0µC -2.05µC

= - 4.05µC (note that the charges on first and second objects are now the same)

Thus, 2.05µC of electrons must be transferred from the first to the second object so that both objects have the same charge

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A solenoidal coil with 21 turns of wire is wound tightly around another coil with 350 turns. The inner solenoid is 22.0 cm long
Thepotemich [5.8K]

Answer:

(a)The average of magnetic flux trough each turns is 3.63 \times 10^{-8} wb.

(b) The magnitude of  mutual inductance of the two solenoid is 1.596 \times 10^{-5} H.

(c)The magnitude of emf induced in the outer solenoid by changing  current in the inner solenoid is 0.027132 v .

Explanation:

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The diameter of the solenoid=D= 2.20 cm=0.022

The length of the solenoid is = l = 22.0 cm=0.22 m

The second solenoid with N_2= 21 turns which is wound around the solenoid at its center.

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(a)

The formula of magnetic field of a solenoid is

B=\mu_0 nI=\frac{\mu_0 N_1I_1}{l}

                 =\frac{(4\pi \times 10^{-7} T.m/A) (350)(0.15 \ A)}{0.220 \ m}

                \approx 3.0 \times 10^{-4} T

So, the magnetic flux of each turns

\phi = BA=B \pi (\frac d2)^2

            =(3.0\times 10^{-4}\ T)\pi (\frac {0.022}{2} \ m)^2

             =1.14 \times 10^{-7}  wb

The average of magnetic flux trough each turns is 3.63 \times 10^{-8} wb.

(b)

Since the both coil wound tightly. So, the magnitude magnetic flux though each turns of both coils is same.

The mutual inductance of the two solenoid is

M=\frac{N_2\phi}{I_1}

     =\frac{(21)(1.14 \times 10^{-7}\ wb)}{0.15\ A}

    =1.596 \times 10^{-5} H

(c)

The formula of emf is

\varepsilon =-M\frac{dI_1}{dt}

  =-(1.596\times 10^{-5}\ H) (1700 \ A/s)

  =27.132 \times 10^{-3} v        

 =0.027132 v

 The emf induced in the outer solenoid by changing  current in the inner solenoid is 0.027132 v .    

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