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In Fig 5.4 (b), the magnetic needle has ...

In Fig 5.4 (b), the magnetic needle has magnetic moment `6.7 xx 10^(-2) Am^(2)` and moment of inertia `I=7.5 xx 10^(-6) Kg m^(-2)`. It performs 10 complete oscillation in 6.70s. What is the magnitude of the magnetic field?

Text Solution

Verified by Experts

When a magnetic dipole with dipole moment m performs angular simple harmonic oscillations in a uniform magnetic field B, its periodic time is,
`T= 2 pi sqrt((I)/( mB))" "…(1)`
where I = moment of inertia of magnetic dipole (here magnetic needle) about axis of rotation (or about suspension wire or about equatorial line).
`therefore T^(2) = 4pi^(2) ((I)/( mB))`
`therefore B= 4pi^(2) ((I)/( mT^(2)))" "...(2)`
If `n` no. of oscillations are completed in time `t` then time taken to complete one oscillation would be `(t)/( n)` which is called periodic time, shown by symbol T. Thus,
`T= (t)/( n) = (6.7)/( 10) = 0.67` s
From equation (2),
`B= (4) (3.14)^(2) ((7.5 xx 10^(-6))/( 6.7 xx 10^(-2) xx (0.67)^(2) ))`
`therefore B= 9.834 xx 10^(-3) T`
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Knowledge Check

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