NEET Ampere's Circuital Law MCQs & PYQs

Practice NEET Ampere's Circuital Law Questions

Question 11:

easy

Assertion (A): A rectangular current loop is in an arbitrary orientation in an external uniform magnetic field. No work is required to rotate the loop about an axis perpendicular to its plane.

Reason (R): All positions represent the same level of energy.

Assertion (A): A current loop in a uniform magnetic field experiences a torque $\vec{\tau} = \vec{M} \times \vec{B}$. Work is generally required to change its orientation. So, (A) is false. Reason (R): The potential energy of a current loop in a magnetic field is $U = -\vec{M} \cdot \vec{B}$, which depends on the orientation of $\vec{M}$ relative to $\vec{B}$. Thus, not all positions represent the same energy. So, (R) is false. Both (A) and (R) are false.

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Question 12:

easy

Assertion (A): In Ampere’s law for magnetostatics $\oint \vec{B} \cdot d\vec{l} = \mu_0 \sum I_{\text{i}}$ the current outside the Amperian loop is not included on the right side.

Reason (R): Magnetic field calculated using Ampere’s law is due to inside as well outside the current of closed loop.

Assertion (A): Ampere's law $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}}$ states that only currents passing through the Amperian loop contribute to the right-hand side. So, (A) is true.

Reason (R): The magnetic field (vec{B}) on the left-hand side of Ampere's law is the total field produced by all currents, both inside and outside the loop. So, (R) is true. However, R describes the nature of (vec{B}), not why only enclosed currents are counted on the right side. Thus, (R) is not the correct explanation of (A).

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Question 13:

easy

A long solenoid carrying a current produces a magnetic field $B$ along its axis. If the current is doubled and the number of turns per cm is halved, the new value of the magnetic field is:

(2003)

The magnetic field of a solenoid is $B = mu_0 n I$. When the current is doubled ($I' = 2I$) and turns per unit length are halved ($n' = n/2$), the new magnetic field is $B' = mu_0 (n/2)(2I) = mu_0 n I = B$.

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Question 14:

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 15:

easy

A long solenoid of radius $1\text{ mm}$ has $100$ turns per mm. If $1\text{ A}$ current flows in the solenoid, the magnetic field strength at the centre of the solenoid is:

(2022)

The magnetic field inside a long solenoid is given by $B = \mu_0 n I$. Given $n = 100\text{ turns/mm} = 10^5\text{ turns/m}$ and $I = 1\text{ A}$, substituting the values yields $B = (4\pi \times 10^{-7}) \times 10^5 \times 1 = 12.56 \times 10^{-2}\text{ T}$.

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Question 16:

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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