In end on and broadside on position of a deflection magnetometer- if -1 and -2 are the deflections produced by short magnets at equal distances- then tan- 1 -tan- 2 is

The correct option is A 2:1 B = B_Htanθ End On, Tan A: μ_02Md4π (d^2 l^2)^2 Tan B: μ_0M4π (d^2+l^2)^3/2 tanθ_1tanθ_2 = B_1:B_2 = 2:1 ...

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Byju's Answer

Standard XII

Physics

Capacitance of Parallel Plate Capacitor

In end on and...

Question

In end on and broadside on position of a deflection magnetometer, if $θ_{1}$ and $θ_{2}$ are the deflections produced by short magnets at equal distances, then $tan θ_{1} / tan θ_{2}$ is

2 : 1

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1 : 2

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None of these

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Solution

The correct option is A $2 : 1$
$B = B_{H} t a n θ$

End On, Tan A: $\frac{μ_{0} 2 M d}{4 π (d^{2} - l^{2})^{2}}$

Tan B: $\frac{μ_{0} M}{4 π (d^{2} + l^{2})^{3 / 2}}$

$\frac{t a n θ_{1}}{t a n θ_{2}} = B_{1} : B_{2} = 2 : 1$ $(l << d)$

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In end on and broadside on position of a deflection magnetometer, if θ1 and θ2 are the deflections produced by short magnets at equal distances, then tanθ1/tanθ2 is

The correct option is A 2:1B=BHtanθEnd On, Tan A: μ02Md4π(d2−l2)2Tan B: μ0M4π(d2+l2)3/2tanθ1tanθ2=B1:B2=2:1 (l<<d)

In end on and broadside on position of a deflection magnetometer, if $θ_{1}$ and $θ_{2}$ are the deflections produced by short magnets at equal distances, then $tan θ_{1} / tan θ_{2}$ is

The correct option is A $2 : 1$
$B = B_{H} t a n θ$

End On, Tan A: $\frac{μ_{0} 2 M d}{4 π (d^{2} - l^{2})^{2}}$

Tan B: $\frac{μ_{0} M}{4 π (d^{2} + l^{2})^{3 / 2}}$

$\frac{t a n θ_{1}}{t a n θ_{2}} = B_{1} : B_{2} = 2 : 1$ $(l << d)$