Match the LIST-I with LIST-II (Symbols in the formulae have their usual meaning)
| LIST-I (Physical Law/Quantity) | LIST-II (Formulae) |
| A. Biot Savart's Law | I. \( \displaystyle\frac{\mu_0 I_1 I_2}{2\pi d} \) |
| B. Ampere's Circuital Law | II. \( q[\vec{E} + \vec{v} \times \vec{B}] \) |
| C. Force between two parallel current carrying conductors | III. \( \oint \vec{B} \cdot d\vec{l} = \mu_0 \Sigma i \) |
| D. Lorentz Force | IV. \( \vec{B} = \displaystyle\frac{\mu_0 i}{4\pi} \textstyle\int \displaystyle\frac{i\,dl\sin\theta}{r^{2}}\,\hat{n} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A - IV, B - III, C - I, D - II
(A) Biot Savart’s Law — this matches (IV). The Biot–Savart law gives the field of a current element; integrating it gives the field of the whole conductor.
\( \displaystyle dB = \frac{\mu_0}{4\pi}\,\frac{i\,dl\sin\theta}{r^{2}} \)
(B) Ampere’s Circuital Law — this matches (III) \( \oint \vec{B} \cdot d\vec{l} = \mu_0 \Sigma i \). The line integral of B around any closed loop equals \( \mu_0 \) times the net current threading the loop.
(C) Force between two parallel current carrying conductors — this matches (I). Each wire lies in the field \( \mu_0 I_1/2\pi d \) of the other; expression (I) is the force per unit length.
\( \displaystyle \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} \)
(D) Lorentz Force — this matches (II) \( q[\vec{E} + \vec{v} \times \vec{B}] \). The Lorentz force is the total force on a charge q moving with velocity v through electric and magnetic fields.
Note: It seems that there is a mistake in the original NTA paper, which prints entry (IV) as \( \vec{B} = \frac{\mu_0 i}{4\pi}\int\frac{i\,dl\sin\theta}{r^{2}}\,\hat{n} \), with the current i written twice, when it should read \( \vec{B} = \frac{\mu_0}{4\pi}\int\frac{i\,dl\sin\theta}{r^{2}}\,\hat{n} \). Also, entry (I) is the force per unit length between the two conductors, not the force itself. The question can still be solved: only Option (4) pairs (B) with (III).
NCERT (2026-27):
Magnetic Field due to a Current Element, Biot-Savart Law – “… The magnitude of this field is,
\( \displaystyle |d\mathbf{B}| = \frac{\mu_0}{4\pi}\,\frac{I\,dl\sin\theta}{r^{2}} \qquad [4.7(b)] \)”
Ampere’s Circuital Law – “… Ampere’s law states that this integral is equal to \( \mu_0 \) times the total current passing through the surface, i.e.,
\( \displaystyle \oint \mathbf{B}\cdot d\mathbf{l} = \mu_0 I \qquad [4.13(a)] \)”
Force Between Two Parallel Currents, the Ampere – “… The magnitude of this force is given by Eq. (4.4),
\( \displaystyle F_{ba} = I_b L B_a = \frac{\mu_0 I_a I_b}{2\pi d}\,L \qquad (4.17) \)”
Magnetic Field, Lorentz Force – “… The force on an electric charge \( q \) due to both of them can be written as
\( \displaystyle \mathbf{F} = q\left[\mathbf{E}(\mathbf{r}) + \mathbf{v}\times\mathbf{B}(\mathbf{r})\right] \equiv \mathbf{F}_{electric} + \mathbf{F}_{magnetic} \qquad (4.3) \)
This force was given first by H.A. Lorentz based on the extensive experiments of Ampere and others. It is called the Lorentz force.”