Match List I with List II

Choose the correct answer from the option given below:

Solution :

For a circular arc carrying current, magnetic field at centre is :

$$B = \frac{\mu_0 I \theta}{4\pi r}$$

For semicircle :

$$\theta = \pi$$

So,

$$B_{semi} = \frac{\mu_0 I}{4r}$$

Magnetic field due to straight wire at distance r :

$$B = \frac{\mu_0 I}{4\pi r}$$

Case A :

Magnetic field due to full circular loop :

$$B_1 = \frac{\mu_0 I}{2r}$$

Field due to straight wire :

$$B_2 = \frac{\mu_0 I}{2\pi r}$$

Both are in same direction.

$$B_0 = \frac{\mu_0 I}{2\pi r}(\pi+1)$$

Hence,

A → IV

Case B :

Magnetic field due to semicircle :

$$B_1 = \frac{\mu_0 I}{4r}$$

Fields due to two straight wires add :

$$B_2 = \frac{\mu_0 I}{2\pi r}$$

Hence,

$$B_0 = \frac{\mu_0 I}{4\pi r}(\pi+2)$$

Therefore,

B → I

Case C :

Magnetic field due to semicircle :

$$B_1 = \frac{\mu_0 I}{4r}$$

Field due to straight wire is opposite in direction.

Hence,

$$B_0 = \frac{\mu_0 I}{2\pi r}(\pi-1)$$

Therefore,

C → III

Case D :

Only semicircular arc contributes significantly.

$$B_0 = \frac{\mu_0 I}{4r}$$

Therefore,

D → II

Final Answer :

A-IV, B-I, C-III, D-II