Question

A bar magnet of mass 120 g, in the form of a rectangular parallelepiped, has dimensions1= 40 mm, b= 10 mm and h= 80 mm. With the dimension h vertical, the magnet performsangular oscillations in the plane of a magnetic field with period i s. If its magnetic momentis 3.4 A m?, determine the influencing magnetic field.​

Answer 1

Answer:

The moment of inertia of the magnet about the axis of rotation is <br>

<br>

<br>

. <br> We have, <br>

… (i) <br> or,

<br> T

...(ii) <br> Dividing by equation (i),

<br>

<br>

<br> or,

Explanation:

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Answer 2

Answer:

The influencing magnetic field for the given condition is of strength $2*10^{-5}$ T.

Explanation:

Given is the information about the question as,

mass of magnet(m)=120g=0.12kg,

l=40mm=0.04m, b=10mm=0.01m, h=80mm=0.08m

Time period(T)=π s

Magnetic Moment(u)=3.4Am²

Step 1:

$I=m(\frac{l^{2} +b^{2} }{12} )$

$I=0.12(\frac{1600 +100}{12} )*10^{-6} \\=1.7*10^{-5} Am^{2}$

Step 2:

$T=2\pi\sqrt{\frac{I}{uB} }\\ \pi =2\pi \sqrt{\frac{I}{uB} }\\Squaring\ both\ the\ sides,\\\pi ^{2} =4\pi ^{2}\frac{I}{uB}\\B=\frac{4I}{u} \\B=\frac{4*1.7*10^{-5} }{3.4} \\B=2*10^{-5} T$

Hence, the influencing magnetic field is of strength $2*10^{-5}$ T.