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A magnetic needle placed in uniform magn...

A magnetic needle placed in uniform magnetic field has magnetic moment `6.7 xx 10^(-2) "Am"^(2)`, and moment of inertia of `15 xx 10^(-6)` k `m^(2)`. It performs 10 complete oscillations in 6.70 s. What is the magnitude of the magnetic field ?

Text Solution

Verified by Experts

Dipole moment ` = 6.7 xx 10^(-2) "Am"^(2)`
Moment of inertia `I= 15 xx 10^(-6) "Kg m"^2`
Periodic time T `= ("time")/("No. of oscillations")`
`- (6.7)/(10) -0.67` s
The period of magnetic needle placed in uniform magnetic field,
`T= 2pi sqrt((I)/( mB))`
`therefore T^(2) = (4 pi^(2) I)/( mB)`
`therefore B= (4pi^(2) I)/( mT^(2) )`
`= (4 xx (3.14)^(2) xx 15 xx 10^(-6) )/( 6.7 xx 10^(-2) xx (0.67)^(2) )`
`= 196.69 xx 10^(-4)`
`therefore B ~~ 0.02` T
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Knowledge Check

  • Force acting on a magnetic pole of 7xx 10^(-2) Am is 31.5 N. Magnetic field at that point is …

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    `4 xx 10^(-2) T`
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    `4.5 xx 10^(-2) T`
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    `3 xx 10^(2) T`

  • The intensity of magnetization of a magnet of magnetic dipole moment 1.2 "Am"^(2) and dimension 0.15 m xx 0.02 m xx 0.01m is…..... A/m.

    A
    `4 xx 10^(4) A//m`
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    `2 xx 10^(4) A//m`
    C
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    D
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