A bar magnet is demagnetized by inserting it inside a solenoid of length 0.2 m, 100 turns, and carrying a curent of 5.2 A. The coercivity of the bar magnet is:

$\text{Given}$:Length of solenoid l = 0.2 m

              No. of turns N = 100

              Current I = 5.2 A

$\text{To find}$ :  Coercivity of bar magnet H

$\text{Solution}$ : 

Coercivity of bar magnet H = nI

 Now, n = $\frac{N}{l}$

Therefore, H = $\frac{N}{l}\times I$ = $\frac{100}{0.2}\times 5.2$ = 26$\times 100$ = 2600

H = 2600 A/m

$\text{Hence A is the correct option}$