A bar magnet of 10 cm long is kept with its north (N)-pole pointing North. A neutral point is formed at a distance of 15 cm from each pole. Given the horizontal component of earth's field is 0.4 Gauss, the pole strength of the magnet is :
Answers
Answer:
Explanation:
=> Magnet's length L = 10 cm = 10⁻¹ m
A neutral point is formed at a distance of 15 cm from each pole.
r = 15 cm = 15 * 10⁻² m
=> PN² = OP² + ON²
OP = √PN² - ON²
= √15² - 5²
= √225 - 25 cm
= √200 cm"
=> According to tangent law,
B_H = (μ₀/4π) [{(2l)(m)}/(OP² + AO²)³/²]
∴0.4 * 10⁻⁴ = [(4π * 10⁻⁷ * 10⁻¹ * m)/ (4π(200 + 25) * 10⁻⁴)³/²]
0.4 * 10⁴ = [m/(15*10⁻²)³]
Thus, m = 13.5 A.m.
Therefore, the pole strength of the magnet is 13.5 A.m.
Explanation:
Magnet's length L = 10 cm = 0.1 m
A neutral point is formed at a distance of 15 cm from each pole.
TIP :- As when we use the eq (a2 + b2) ^3/2 Which is actually this (15)^3
try to skip the below step
r = 15 cm = 0.15m
OP = √PN² - ON²
= √15² - 5²
= √225 - 25 cm = √200 cm
BH = (μ₀/4π) [P/(OP² + AO²)³/²]
∴0.4 * 10⁻⁴ = [(10⁻⁷ ) P/ (15)³]
0.4 * 10⁴ = [m/(15*10⁻²)³]
Thus, P = 13.5 A.m.
Hope you understand
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