Magnetic dipole in a uniform magnetic field

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Question

At a point on the surface of earth, time period of oscillation of magnetic needle is T. Now a uniform magnetic field B, perpendicular to earth's horizontal component of magnetic field is produced in that region and time period of oscillation of the same needle at the same place is found to be $\frac{T}{\sqrt{3}}$ . Then $\frac{B}{B_{H}}$ is equal to

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Solution

$T = 2 π \sqrt{\frac{I}{{MB}_{H}}} \Rightarrow B_{H} = \frac{4 π^{2} I}{{MT}^{2}}$
$T_{1} = 2 π \sqrt{\frac{I}{{MB}_{1}}} \Rightarrow B_{1} = \frac{4 π^{2} I}{{MT}_{1}^{2}} = \frac{B_{H} T^{2}}{{T_{1}}^{2}}$
$Now, \sqrt{B^{2} + {B_{H}}^{2}} = B_{1}$
$Given T_{1} = \frac{T}{\sqrt{3}}$
$∴ \sqrt{B^{2} + {B_{H}}^{2}} = B_{H} {(\frac{T}{T_{1}})}^{2} = B_{H} {(\sqrt{3})}^{2} = 3 B_{H}$
$\Rightarrow \frac{B}{B_{H}} = 2 \sqrt{2}$

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Magnetic dipole in a uniform magnetic field

Remember concepts with our Masterclasses.

T=2πIMBH⇒BH=4π2IMT2T1=2πIMB1⇒B1=4π2IMT12=BHT2T12Now, B2+BH2=B1 Given T1=T3∴B2+BH2=BHTT12=BH32=3BH ⇒BBH=22

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