Match List-I with List-II (Symbols have their usual meaning)List-ILis…

The correct answer is Option (4) → (A) - (III), (B) - (IV), (C) - (I), (D) - (II)

Each entry is a standard result of moving charges and magnetism.

(A) Magnetic field due to an infinitely long straight conductor — this matches (III) \( \frac{\mu_0 I}{2\pi r} \). By Ampere’s circuital law, the field lines are circles round the wire and \( B(2\pi r) = \mu_0 I \).

(B) Magnetic field due to a circular coil at the center — this matches (IV) \( \frac{\mu_0 I}{2r} \). By the Biot–Savart law, every element of the loop is at distance r from the centre and perpendicular to the radius, so the contributions add to \( \frac{\mu_0 I}{4\pi r^{2}} \times 2\pi r \).

(C) Force on a current carrying conductor in a magnetic field — this matches (I) BIl sinθ. The force \( \vec{F} = I\,\vec{l} \times \vec{B} \) has magnitude \( BIl\sin\theta \), where θ is the angle between the conductor and the field.

(D) Torque on a current loop in a magnetic field — this matches (II) BIA sinθ. A loop of area A has magnetic moment m = IA, and the torque is \( \tau = mB\sin\theta = BIA\sin\theta \) (for a single turn).

NCERT (2026-27):
Ampere’s Circuital Law – “… We find that the magnetic field at a distance r outside the wire is tangential and given by

    \( \displaystyle B \times 2\pi r = \mu_0 I, \quad B = \mu_0 I/(2\pi r) \qquad (4.14) \)”

Magnetic Field on the Axis of a Circular Current Loop – “… As a special case of the above result, we may obtain the field at the centre of the loop. Here x = 0, and we obtain,

    \( \displaystyle \mathbf{B}_0 = \frac{\mu_0 I}{2R}\,\hat{\mathbf{i}} \qquad (4.12) \)”

Magnetic Force on a Current-Carrying Conductor – “… Thus,

    \( \displaystyle \mathbf{F} = [(nq\,\mathbf{v}_d)lA] \times \mathbf{B} = [jAl] \times \mathbf{B} = I\,\mathbf{l} \times \mathbf{B} \qquad (4.4) \)

where l is a vector of magnitude l, the length of the rod, and with a direction identical to the current I.”

Torque on a Rectangular Current Loop in a Uniform Magnetic Field – “… The magnitude of the torque on the loop is,

    \( \displaystyle \tau = F_1\frac{a}{2}\sin\theta + F_2\frac{a}{2}\sin\theta = I ab B \sin\theta = I A B \sin\theta \qquad (4.21) \)”