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Iteru [2.4K]
3 years ago
14

Consider first the generation of the magnetic field by the current I1(t)I1(t)I_{1}(t) in solenoid

Physics
1 answer:
densk [106]3 years ago
7 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

So the magnetic field on the solenoid 1 is would be

                 B_1 (t) = \mu_0 n_1 I_1 (t)

Explanation:

Now in this question we are given an solenoid and our interest is on the magnetic field on the solenoid

Now first we will consider the Ampere's law which is mathematically represented as

                        ∮\= B \cdot d \=s = \mu_0 I_{en}

 Where \=B is the magnetic filed

             ∮ means the close integral

             d\= s  is the length element

              \mu_0 magnetic of permeability of free space

               I_{en} is the enclosed current

Now choosing a current path on a  particular side of the solenoid (as shown on the second uploaded image)  to evaluate the magnetic field

     Where I is the current elements within the enclosed current path

                 N is the number of current elements in that particular enclosed current path.

          and L is the length of enclosed path in the direction of the magnetic field as shown on the second uploaded image

 

So the integral of total  path  ∮\= B \cdot d \=s = \int\limits { \= B } \, d\=s_1 + \int\limits { \= B } \, d\=s_2 +\int\limits { \= B } \, d\=s_3 +\int\limits { \= B } \, d\=s_4

Where s_1 ,s_2 ,s_3,s_4 are the length element of each sides of the enclosed path

   Now looking at the first integral we see that s_1 is moving the same direction with magnetic field B  as shown on the third uploaded image  

 Now for s_2 the direction is perpendicular to the the direction of the magnetic field B so the value of magnetic field for that segment is 0 \

Now for s_3 from the third uploaded image we see that there are no this segment so the value is zero

s_4 segment is the same as s_2 segment so the value of magnetic field on this segment is zero

   Therefore      

                    ∮\= B \cdot d \=s  = BL + 0 + 0 + 0              

 Now from Ampere's law  ∮\= B \cdot d \=s = \mu_0 I_{en} \equiv BL = \mu_0 I_{en}

Now from the second uploaded image we see that I_{en} would now be equal to NI

               Therefore the equation becomes

                   BL = \mu_0 (NI)

Now making B the subject we have

                  B = \frac{\mu_o NI}{L}  = \mu_0 \frac{N}{L} I

Now  \frac{N}{L} = n where n is the number of turns per length  

                    So we have

                            B = \mu_0 n I

So the magnetic field on the solenoid 1 is would be

           B_1 (t) = \mu_0 n_1 I_1 (t)

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Answer:

0.075 m

Explanation:

The picture of the problem is missing: find it in attachment.

At first, block A is released at a distance of

h = 0.75 m

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m_Agh=\frac{1}{2}m_Av_A^2

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v_A is the speed of the block A just before touching block B

Solving for the speed,

v_A=\sqrt{2gh}=\sqrt{2(9.8)(0.75)}=3.83 m/s

Then, block A collides with block B. The coefficient of restitution in the collision is given by:

e=\frac{v'_B-v'_A}{v_A-v_B}

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e = 0.7 is the coefficient of restitution in this case

v_B' is the final velocity of block B

v_A' is the final velocity of block A

v_A=3.83 m/s

v_B=0 is the initial velocity of block B

Solving,

v_B'-v_A'=e(v_A-v_B)=0.7(3.83)=2.68 m/s

Re-arranging it,

v_A'=v_B'-2.68 (1)

Also, the total momentum must be conserved, so we can write:

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A wire with a length of 150 m and a radius of 0.15 mm carries a current with a uniform current density of 2.8 x 10^7A/m^2. The c
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Answer:

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(A) is correct option.

Explanation:

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Length = 150 m

Radius = 0.15 mm

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We need to calculate the current

Using formula of current density

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Where, J = current density

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Put the value into the formula

I=2.8\times10^{7}\times\pi\times(0.15\times10^{-3})^2

I=1.97=2.0\ A

Hence, The current is 2.0 A.

7 0
3 years ago
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