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:
2600 A/m
520 A/m
285 A/m
1200 A/m
A
2600 A/m
B
285 A/m
C
520 A/m
D
1200 A/m
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Solution
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Given:Length of solenoid l = 0.2 m
No. of turns N = 100
Current I = 5.2 A
To find : Coercivity of bar magnet H
Solution :
Coercivity of bar magnet H = nI
Now, n = Nl
Therefore, H = Nl×I = 1000.2×5.2 = 26×100 = 2600
H = 2600 A/m
Hence A is the correct option
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