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`         A short magnet produces a deflection of 37° in a deflection magnetometer in Tan-A position when placed at a separation of 10 cm from the needle. Find the ratio of the magnetic moment of the magnet to the earth’s horizontal magnetic field.`
3 years ago

## Answers : (1)

Kevin Nash
332 Points
```										Sol. M/B base H (found in the previous problem) = 3.75 *10^3 A-m^2 T^-1
θ = 37°, d = ?
M/B base H = 4π/μ base 0 (d^2 - ℓ^2)^3/2 tan θ
ℓ << d neglecting ℓ w.r.t.d
⇒ M/B base H = 4π/μ base 0 d^3 Tanθ⇒ 3.75 * 10^3 = 1/10^-7 * d^3 * 0.75
⇒ d^3 = 3.75 * 10^3 * 10^-7/0.75 = 5 * 10^-4
⇒ d = 0.079 m = 7.9 cm

```
3 years ago
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