Ampere's Circuital Law & Solenoid — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Ampere's Circuital Law & Solenoid MCQs with step-by-step solutions (21 questions). Part of Magnetic Fields due to Electric Current. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Ampere's Circuital Law & Solenoid · easy · theory
Ampere's circuital law states that:
A. $\oint \vec{B}\cdot d\vec{l} = \mu_0I_{enc}$ ✓ Correct
B. $\oint \vec{B}\cdot d\vec{A} = \mu_0I_{enc}$
C. $\oint \vec{B}\times d\vec{l} = \mu_0I_{enc}$
D. $\oint \vec{B}\cdot d\vec{l} = 0$ always
Solution: The line integral of $\vec{B}$ round any closed path equals $\mu_0$ times the net current threading that path.
Q2 — Ampere's Circuital Law & Solenoid · easy · theory
The magnetic field inside a long solenoid with $n$ turns per unit length carrying current $I$ is:
A. $\dfrac{\mu_0nI}{2}$
B. $\dfrac{\mu_0I}{2n}$
C. $\mu_0n^2I$
D. $\mu_0nI$ ✓ Correct
Solution: Applying Ampere's law to a rectangular loop partly inside the solenoid gives this uniform interior field.
Q3 — Ampere's Circuital Law & Solenoid · medium · theory
The magnetic field inside a long solenoid is:
A. Directed radially outward
B. Zero along the axis
C. Uniform and parallel to the axis ✓ Correct
D. Strongest near the walls
Solution: Away from the ends the field lines run straight and evenly spaced, which is why solenoids are used to produce uniform fields.
Q4 — Ampere's Circuital Law & Solenoid · medium · theory
The magnetic field at the end of a long solenoid compared with that at its centre is:
A. Equal
B. Zero
C. Half as large ✓ Correct
D. Twice as large
Solution: At the end only half the solenoid contributes on each side, giving $B = \dfrac{\mu_0nI}{2}$.
Q5 — Ampere's Circuital Law & Solenoid · medium · theory
The magnetic field inside a toroid of $N$ turns and mean radius $r$ carrying current $I$ is:
A. $\dfrac{\mu_0NI}{2\pi r}$ ✓ Correct
B. $\mu_0NI$
C. $\dfrac{\mu_0NI}{4\pi r}$
D. $\dfrac{\mu_0NI}{2r}$
Solution: A circular Amperian loop inside the winding encloses $NI$, giving a field that varies inversely with $r$.
Q6 — Ampere's Circuital Law & Solenoid · medium · theory
The magnetic field outside an ideal long solenoid is:
A. Equal to the internal field
B. Practically zero ✓ Correct
C. Radially outward
D. Twice the internal field
Solution: The return flux spreads over a very large area, so the external field is negligible compared with the interior one.
Q7 — Ampere's Circuital Law & Solenoid · easy · theory
For a solenoid of $N$ turns wound over a length $L$, the number of turns per unit length is:
A. $\dfrac{N}{2L}$
B. $NL$
C. $\dfrac{N}{L}$ ✓ Correct
D. $\dfrac{L}{N}$
Solution: This quantity $n$ is what enters the expression $B = \mu_0nI$.
Q8 — Ampere's Circuital Law & Solenoid · medium · theory
Ampere's circuital law is most useful for calculating fields in situations possessing:
A. A high degree of symmetry ✓ Correct
B. Only electrostatic charges
C. Only time-varying currents
D. No symmetry at all
Solution: Symmetry lets $B$ be taken outside the integral, just as symmetry makes Gauss' law practical in electrostatics.
Q9 — Ampere's Circuital Law & Solenoid · medium · numerical
A solenoid has $1000$ turns per metre and carries $2\text{ A}$. The field inside it is approximately:
A. $1.26 \times 10^{-3}\text{ T}$
B. $5 \times 10^{-3}\text{ T}$
C. $2.5 \times 10^{-4}\text{ T}$
D. $2.5 \times 10^{-3}\text{ T}$ ✓ Correct
Solution: $B = \mu_0nI = 4\pi \times 10^{-7} \times 1000 \times 2 \approx 2.51 \times 10^{-3}\text{ T}$.
Q10 — Ampere's Circuital Law & Solenoid · hard · numerical
A solenoid of $500$ turns wound over $0.5\text{ m}$ carries $1\text{ A}$. The field inside is approximately:
A. $2.5 \times 10^{-3}\text{ T}$
B. $1.26 \times 10^{-4}\text{ T}$
C. $6.3 \times 10^{-4}\text{ T}$
D. $1.26 \times 10^{-3}\text{ T}$ ✓ Correct
Solution: $n = \dfrac{500}{0.5} = 1000\text{ m}^{-1}$, so $B = 4\pi \times 10^{-7} \times 1000 \times 1 \approx 1.26 \times 10^{-3}\text{ T}$.
Q11 — Ampere's Circuital Law & Solenoid · medium · numerical
The field at the centre of a solenoid is $1.26 \times 10^{-3}\text{ T}$. The field at one of its ends is:
A. $2.5 \times 10^{-3}\text{ T}$
B. Zero
C. $1.26 \times 10^{-3}\text{ T}$
D. $6.3 \times 10^{-4}\text{ T}$ ✓ Correct
Solution: The end field is exactly half the central value.
Q12 — Ampere's Circuital Law & Solenoid · hard · numerical
A toroid of $500$ turns has a mean radius of $0.1\text{ m}$ and carries $2\text{ A}$. The field inside is approximately:
A. $1.26 \times 10^{-3}\text{ T}$
B. $2 \times 10^{-4}\text{ T}$
C. $2 \times 10^{-3}\text{ T}$ ✓ Correct
D. $4 \times 10^{-3}\text{ T}$
Solution: $B = \dfrac{\mu_0NI}{2\pi r} = \dfrac{4\pi \times 10^{-7} \times 500 \times 2}{2\pi \times 0.1} = 2 \times 10^{-3}\text{ T}$.
Q13 — Ampere's Circuital Law & Solenoid · easy · numerical
If the number of turns per unit length of a solenoid is doubled at constant current, the interior field:
A. Halves
B. Becomes four times
C. Remains unchanged
D. Doubles ✓ Correct
Solution: $B = \mu_0nI \propto n$.
Q14 — Ampere's Circuital Law & Solenoid · easy · numerical
If the current through a solenoid is doubled, the field inside it:
A. Becomes four times
B. Remains unchanged
C. Halves
D. Doubles ✓ Correct
Solution: $B = \mu_0nI \propto I$.
Q15 — Ampere's Circuital Law & Solenoid · hard · numerical
A closed Amperian loop encloses a net current of $2\text{ A}$. The value of $\oint \vec{B}\cdot d\vec{l}$ is:
A. $2 \times 10^{-7}\text{ T}\cdot\text{m}$
B. Zero
C. $2.51 \times 10^{-6}\text{ T}\cdot\text{m}$ ✓ Correct
D. $1.26 \times 10^{-6}\text{ T}\cdot\text{m}$
Solution: $\oint \vec{B}\cdot d\vec{l} = \mu_0I = 4\pi \times 10^{-7} \times 2 \approx 2.51 \times 10^{-6}\text{ T}\cdot\text{m}$.
Q16 — Ampere's Circuital Law & Solenoid · medium · numerical
A solenoid has $2000$ turns per metre and carries $0.5\text{ A}$. The field inside is approximately:
A. $1.26 \times 10^{-4}\text{ T}$
B. $2.5 \times 10^{-3}\text{ T}$
C. $6.3 \times 10^{-4}\text{ T}$
D. $1.26 \times 10^{-3}\text{ T}$ ✓ Correct
Solution: $B = 4\pi \times 10^{-7} \times 2000 \times 0.5 \approx 1.26 \times 10^{-3}\text{ T}$.
Q17 — Ampere's Circuital Law & Solenoid · hard · numerical
A solenoid is stretched to twice its length while the number of turns stays the same. The interior field:
A. Becomes one-fourth
B. Halves ✓ Correct
C. Doubles
D. Remains unchanged
Solution: Stretching halves $n = \dfrac{N}{L}$, and $B \propto n$, so the field halves.
Q18 — Ampere's Circuital Law & Solenoid · medium · numerical
A solenoid is filled with a core of relative permeability $1000$. The field inside becomes:
A. Unchanged
B. $1000$ times smaller
C. $1000$ times as large ✓ Correct
D. $\sqrt{1000}$ times as large
Solution: With a magnetic core, $B = \mu_r\mu_0nI$, so the field is multiplied by the relative permeability.
Q19 — Ampere's Circuital Law & Solenoid · medium · numerical
The magnetic field deep inside a long solenoid depends on:
A. The radius of the solenoid
B. The distance from the axis
C. The turns per unit length and the current only ✓ Correct
D. The total length of the solenoid
Solution: $B = \mu_0nI$ contains neither the radius nor the position within the cross-section.
Q20 — Ampere's Circuital Law & Solenoid · medium · numerical
A solenoid has $5000$ turns per metre and carries $0.2\text{ A}$. The field inside is approximately:
A. $6.3 \times 10^{-4}\text{ T}$
B. $1.26 \times 10^{-3}\text{ T}$ ✓ Correct
C. $1.26 \times 10^{-2}\text{ T}$
D. $2.5 \times 10^{-3}\text{ T}$
Solution: $B = 4\pi \times 10^{-7} \times 5000 \times 0.2 \approx 1.26 \times 10^{-3}\text{ T}$.
Q21 — Ampere's Circuital Law & Solenoid · medium · numerical
A closed Amperian loop encloses no net current. The value of $\oint \vec{B}\cdot d\vec{l}$ around it is:
A. Infinite
B. $\mu_0$
C. Zero ✓ Correct
D. Equal to $B$ times the perimeter
Solution: Ampere's law gives zero, even though the field at individual points on the loop need not be zero.