Solution:

To solve this problem, we use the formula for the magnetic field inside a long solenoid:

$$B = \mu_0 n i,$$

where:

• $B$ is the magnetic field at the center of the solenoid,
• $\mu_0 = 4\pi \times 10^{-7} \, \text{T·m/A}$ is the permeability of free space,
• $n$ is the number of turns per unit length (in meters),
• $i$ is the current in the solenoid.

Step 1: Magnetic Field for the First Solenoid

The first solenoid has:

• $n_1 = 200 \, \text{turns per cm} = 200 \times 100 = 20000 \, \text{turns per meter}$,
• $i_1 = i$,
• $B_1 = 6.28 \times 10^{-2} \, \text{weber/m}^2$.

Using the formula for $B$, substitute $B_1$ to find $i$:

$$B_1 = \mu_0 n_1 i,$$ $$6.28 \times 10^{-2} = (4\pi \times 10^{-7}) \cdot 20000 \cdot i.$$

Simplify:

$$i = \frac{6.28 \times 10^{-2}}{4\pi \times 10^{-7} \cdot 20000}.$$

Substitute $4\pi \approx 12.57$:

$$i = \frac{6.28 \times 10^{-2}}{12.57 \times 10^{-7} \cdot 20000} = \frac{6.28 \times 10^{-2}}{2.514 \times 10^{-2}}.$$ $$i = 2.5 \, \text{A}.$$

Step 2: Magnetic Field for the Second Solenoid

The second solenoid has:

• $n_2 = 100 \, \text{turns per cm} = 100 \times 100 = 10000 \, \text{turns per meter}$,
• $i_2 = \frac{i}{3} = \frac{2.5}{3} = 0.833 \, \text{A}$.

Using the formula for $B$:

$$B_2 = \mu_0 n_2 i_2,$$

Substitute the values:

$$B_2 = (4\pi \times 10^{-7}) \cdot 10000 \cdot 0.833.$$

Simplify:

$$B_2 = 4\pi \times 10^{-7} \cdot 8330.$$

Substitute $4\pi \approx 12.57$:

$$B_2 = 12.57 \times 10^{-7} \cdot 8330 = 1.048 \times 10^{-2}.$$

Final Answer:

$$\boxed{B_2 = 1.05 \times 10^{-2} \, \text{weber/m}^2}$$

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Solution:

To calculate the magnetic field $B$ at a point $P$ within the outer shell of a coaxial cable ($b < r < c$), carrying equal and opposite currents $i$ on the inner and outer conductors, we use Ampère's law and superposition principles.

1. Current Distribution in the Outer Shell

The outer shell carries current $-i$, distributed uniformly across the cross-sectional area of the shell between radii $b$ and $c$.

The current density $J$ in the shell is:

$$J = \frac{-i}{\pi (c^2 - b^2)}.$$

The current enclosed within a radius $r$ ($b < r < c$) in the outer shell is the current $-i$ contributed by the region from $b$ to $r$:

$$I_{\text{enc,shell}} = J \cdot \text{area of shell portion} = J \cdot \pi (r^2 - b^2).$$

Substituting $J$:

$$I_{\text{enc,shell}} = \frac{-i}{\pi (c^2 - b^2)} \cdot \pi (r^2 - b^2) = \frac{-i (r^2 - b^2)}{c^2 - b^2}.$$

2. Net Enclosed Current at Radius $r$

At any point $P$ within the shell ($b < r < c$), the net current enclosed by a loop of radius $r$ is:

$$I_{\text{enc}} = \text{current from the inner wire} + \text{current from the outer shell}.$$

The inner wire contributes $+i$, and the shell contributes $I_{\text{enc,shell}}$:

$$I_{\text{enc}} = i + \frac{-i (r^2 - b^2)}{c^2 - b^2}.$$

Simplify:

$$I_{\text{enc}} = i \left(1 - \frac{r^2 - b^2}{c^2 - b^2}\right).$$

Factorize:

$$I_{\text{enc}} = i \cdot \frac{c^2 - r^2}{c^2 - b^2}.$$

3. Magnetic Field at Radius $r$

Using Ampère's law, the magnetic field $B$ at radius $r$ is:

$$B \cdot 2\pi r = \mu_0 I_{\text{enc}}.$$

Substitute $I_{\text{enc}}$:

$$B \cdot 2\pi r = \mu_0 \cdot i \cdot \frac{c^2 - r^2}{c^2 - b^2}.$$

Solve for $B$:

$$B = \frac{\mu_0 i}{2\pi r} \cdot \frac{c^2 - r^2}{c^2 - b^2}.$$

Final Answer:

$$\boxed{B = \frac{\mu_0 i}{2\pi r} \cdot \frac{c^2 - r^2}{c^2 - b^2}}$$

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Solution:

Inside the pipe current enclosed will be zero so magnetic field will also be zero. Outside the pipe there will be net Magnetic field

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Solution:

As outside the conductor net current enclosed is zero. using ampere circuital law Magnetic field is also zero.

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Solution:

\[ B= \mu_{0}ni \] where n is number of turns per unit length

n= 100/0.1=1000

i=0.5

substituting we get \[ B = 6.28\times 10^{-4} T\]

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Solution:

As Current enclosed is zero.  \[\int_{P}^{}\overrightarrow{B}.\overrightarrow{dl}=0\]

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Solution:

Current enclosed in A = 3i -3i =0

Current enclosed in B = 2i-i=i

Current enclosed in C = - 2i + i= - i

Current enclosed in D= -i

so Correct order of integral of B.dl is B > A > C = D

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Solution:

current enclosed is (1+2+3-1-4)= 1A

using ampere circuital law we get the answer.

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Solution:

To solve for the magnetic field at a distance of $\frac{5R}{4}$ from the axis of a hollow cylindrical wire carrying current $I$, with inner radius $R$ and outer radius $2R$, we use Ampère's Law. The key steps are:

1. Magnetic Field Inside a Hollow Cylinder

For $R < r < 2R$ (points within the shell of the cylinder):

• Current density $J$ is uniform, given by:

  $$J = \frac{I}{\pi \left((2R)^2 - R^2\right)} = \frac{I}{3\pi R^2}.$$

• The current enclosed within a radius $r$ (where $R < r < 2R$) is:

  $$I_{\text{enc}} = J \cdot \text{area enclosed} = J \cdot \pi (r^2 - R^2).$$Substituting $J$:

  $$I_{\text{enc}} = \frac{I}{3\pi R^2} \cdot \pi (r^2 - R^2) = \frac{I}{3R^2}(r^2 - R^2).$$

• Using Ampère's Law:

  $$B \cdot 2\pi r = \mu_0 I_{\text{enc}},$$ $$B = \frac{\mu_0}{2\pi r} \cdot \frac{I}{3R^2}(r^2 - R^2).$$

2. Magnetic Field at $r = \frac{5R}{4}$

Since $\frac{5R}{4}$ lies within the shell ($R < r < 2R$), substitute $r = \frac{5R}{4}$ into the above equation:

$$B = \frac{\mu_0}{2\pi \cdot \frac{5R}{4}} \cdot \frac{I}{3R^2}\left(\left(\frac{5R}{4}\right)^2 - R^2\right).$$

Simplify:

• $r^2 = \left(\frac{5R}{4}\right)^2 = \frac{25R^2}{16}$,
• $r^2 - R^2 = \frac{25R^2}{16} - R^2 = \frac{25R^2}{16} - \frac{16R^2}{16} = \frac{9R^2}{16}$.

Thus:

$$B = \frac{\mu_0}{2\pi \cdot \frac{5R}{4}} \cdot \frac{I}{3R^2} \cdot \frac{9R^2}{16}.$$

Simplify further:

$$B = \frac{\mu_0}{2\pi} \cdot \frac{4}{5R} \cdot \frac{9I}{48} = \frac{\mu_0 I}{2\pi} \cdot \frac{36}{240R}.$$

Final simplification:

$$B = \frac{3}{40} \frac{\mu_0 I}{\pi R}.$$

Final Answer:

$$\boxed{B = \frac{3}{40} \frac{\mu_0 I}{\pi R}}$$

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