NEET Ampere's Circuital Law MCQs & PYQs

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:

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

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:

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

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

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

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

n= 100/0.1=1000

i=0.5

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

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

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

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

using ampere circuital law we get the answer.

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:

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

Inside the hollow pipe, the magnetic field is zero according to Ampere's law. Since a current-carrying pipe is electrically neutral, the electric field outside the pipe is zero.