Match List-I with List-II (Symbols have their usual meaning)
| List-I | List-II |
| Magnetic field | Expression |
| (A) Magnetic field at centre due to a circular current loop of radius r | (I) \( \displaystyle\frac{2\mu_0\pi I}{4\pi r} \) |
| (B) Magnetic field due to a straight infinite current carrying wire at distance r from the wire | (II) \( \displaystyle\frac{\mu_0 I}{4r} \) |
| (C) Magnetic field at centre due to a current carrying semicircular arc | (III) \( \mu_0 nI \) |
| (D) Magnetic field at centre due to a current carrying long solenoid. | (IV) \( \displaystyle\frac{2\mu_0 I}{4\pi r} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (I), (B) - (IV), (C) - (II), (D) - (III)
Write each standard result and bring it to the List-II form, with \( 4\pi \) in the denominator.
(A) Magnetic field at centre due to a circular current loop of radius r — this matches (I), the field at the centre of a circular loop.
\( \displaystyle B = \frac{\mu_0 I}{2r} = \frac{2\mu_0\pi I}{4\pi r} \)
(B) Magnetic field due to a straight infinite current carrying wire at distance r from the wire — this matches (IV), the field of a long straight wire.
\( \displaystyle B = \frac{\mu_0 I}{2\pi r} = \frac{2\mu_0 I}{4\pi r} \)
(C) Magnetic field at centre due to a current carrying semicircular arc — this matches (II): a semicircle gives half the field of a full loop.
\( \displaystyle B = \frac{1}{2} \times \frac{\mu_0 I}{2r} = \frac{\mu_0 I}{4r} \)
(D) Magnetic field at centre due to a current carrying long solenoid. — this matches (III) \( \mu_0 nI \), with n the number of turns per unit length.
Note: The paper does not state the radius of the semicircular arc in (C); taking it as r, the symbol used in List-II, the question can still be solved.
NCERT (2026-27):
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 Field on the Axis of a Circular Current Loop – “… Hence direction of B for a semicircular arc is given by the right-hand rule and magnitude is half that of a circular loop.”
Ampere’s Circuital Law – “… We find that the magnetic field at a distance r outside the wire is tangential and given by
\( \displaystyle B \times 2\pi r = \mu_0 I, \qquad B = \mu_0 I/(2\pi r) \qquad (4.14) \)”
Summary – “8. The magnitude of the field B inside a long solenoid carrying a current I is \( B = \mu_0 nI \) where n is the number of turns per unit length.”