The coercivity of a small bar magnet is 4× 103 A m- 1 . It is inserted inside a solenoid of 500 turns and length 1 m to demagnetize it. The amount of current to be passed through the solenoid will be

Question

The coercivity of a small bar magnet is 4× 103 A m- 1 . It is inserted inside a solenoid of 500 turns and length 1 m to demagnetize it. The amount of current to be passed through the solenoid will be (A) 2.5 A (B) 5 A (C) 8 A (D) 10 A

Tardigrade Admin · Accepted Answer

n = number of turns per unit length of solenoid , n=(500/1) H=ni⇒ i=(H/n)=(4 × 103/500)=8 A

Q. The coercivity of a small bar magnet is $4 \times 1 0^{3} A m^{- 1}$ . It is inserted inside a solenoid of $500$ turns and length $1 m$ to demagnetize it. The amount of current to be passed through the solenoid will be

$2.5 A$

$5 A$

$8 A$

$10 A$

$n$ = number of turns per unit length of solenoid ,
$n = \frac{500}{1}$
$H = ni \Rightarrow i = \frac{H}{n} = \frac{4 \times 1 0 ^{3}}{500} = 8 A$

Q. The coercivity of a small bar magnet is 4×103Am−1 . It is inserted inside a solenoid of 500 turns and length 1m to demagnetize it. The amount of current to be passed through the solenoid will be

2.5A

5A

8A

10A