Ampere's Circuital Law - NEET Physics Chapterwise MCQs & PYQs

NEET Ampere's Circuital Law MCQs & PYQs

Practice NEET Ampere's Circuital Law Questions

Question 1:

moderate

A long solenoid has 200 turns per cm and carries a current i. The magnetic field at its centre is 6.28 × 10–² weber/m². Another long solenoid has 100 turns per cm and it carries a current i/3. The value of the magnetic field at its centre is

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

 

B=μ0ni,B = \mu_0 n i,

 

where:


  • BB     

    is the magnetic field at the center of the solenoid,


  • μ0=4π×10−7 T\cdotpm/A\mu_0 = 4\pi \times 10^{-7} \, \text{T·m/A}     

    is the permeability of free space,


  • nn     

    is the number of turns per unit length (in meters),


  • ii     

    is the current in the solenoid.

Step 1: Magnetic Field for the First Solenoid

The first solenoid has:


  • n1=200 turns per cm=200×100=20000 turns per metern_1 = 200 \, \text{turns per cm} = 200 \times 100 = 20000 \, \text{turns per meter}     

    ,


  • i1=ii_1 = i     

    ,


  • B1=6.28×10−2 weber/m2B_1 = 6.28 \times 10^{-2} \, \text{weber/m}^2     

    .

Using the formula for

BB

, substitute

B1B_1

to find

ii

:

 

B1=μ0n1i,B_1 = \mu_0 n_1 i,

 

6.28×10−2=(4π×10−7)⋅20000⋅i.6.28 \times 10^{-2} = (4\pi \times 10^{-7}) \cdot 20000 \cdot i.

 

Simplify:

 

i=6.28×10−24π×10−7⋅20000.i = \frac{6.28 \times 10^{-2}}{4\pi \times 10^{-7} \cdot 20000}.

 

Substitute

4π≈12.574\pi \approx 12.57

:

 

i=6.28×10−212.57×10−7⋅20000=6.28×10−22.514×10−2.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 A.i = 2.5 \, \text{A}.

 

Step 2: Magnetic Field for the Second Solenoid

The second solenoid has:


  • n2=100 turns per cm=100×100=10000 turns per metern_2 = 100 \, \text{turns per cm} = 100 \times 100 = 10000 \, \text{turns per meter}     

    ,


  • i2=i3=2.53=0.833 Ai_2 = \frac{i}{3} = \frac{2.5}{3} = 0.833 \, \text{A}     

    .

Using the formula for

BB

:

 

B2=μ0n2i2,B_2 = \mu_0 n_2 i_2,

 

Substitute the values:

 

B2=(4π×10−7)⋅10000⋅0.833.B_2 = (4\pi \times 10^{-7}) \cdot 10000 \cdot 0.833.

 

Simplify:

 

B2=4π×10−7⋅8330.B_2 = 4\pi \times 10^{-7} \cdot 8330.

 

Substitute

4π≈12.574\pi \approx 12.57

:

 

B2=12.57×10−7⋅8330=1.048×10−2.B_2 = 12.57 \times 10^{-7} \cdot 8330 = 1.048 \times 10^{-2}.

 

Final Answer:

 

B2=1.05×10−2 weber/m2\boxed{B_2 = 1.05 \times 10^{-2} \, \text{weber/m}^2}

 

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Question 2:

moderate

In the adjacent figure is shown a closed path P. A long straight conductor carrying a current I passes through O and perpendicular to the plane of the paper. Then which of the following holds good ?

 

 

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

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Question 3:

moderate

Consider six wires coming into or out of the page, all with the same current. Rank the line integral of the magnetic field (from most positive to most negative) taken counter-clockwise around each

loop shown.

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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Question 4:

moderate

Six wires of current I1 = 1A, I2 = 2A, I3 = 3A, I4 = 1A, I5 = 4A, I6 = 5A cut the page perpendicular at points 1,2,3,4,5 and 6. The value of time integral of \(\overrightarrow{B}\) around the dotted closed path \(\left( i.e, \oint_{}^{}\overrightarrow{B} \overrightarrow{dl} \right)\)  is :

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

using ampere circuital law we get the answer.

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Question 5:

moderate

From Ampere’s circuital law for a long straight wire of circular cross section carrying a steady current, the variation of magnetic field in the inside and outside region of the wire is:

(2022)

Inside a uniform cylindrical wire, the magnetic field is directly proportional to the radius ($B \propto r$), increasing linearly. Outside the wire, the magnetic field is inversely proportional to the distance ($B \propto 1/r$). Therefore, option D is the correct choice.

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Question 6:

moderate

A long solenoid of $50text{ cm}$ length having $100$ turns carries a current of $2.5text{ A}$. The magnetic field at the centre of the solenoid is:  ($\mu_0 = 4\pi \times 10^{-7}\text{ T m A}^{-1}$)

(2020)

The number of turns per unit length is $n = \frac{N}{L} = \frac{100}{0.5\text{ m}} = 200\text{ turns/m}$. Using the formula $B = \mu_0 n I$, we substitute $\mu_0 = 4\pi \times 10^{-7}$, $n = 200$, and $I = 2.5\text{ A}$ to get $B = 6.28 \times 10^{-4}\text{ T}$.

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