A bar magnet is oscillating in the Earth's magnetic field with a period T. What happens to its period and motion if its mass is quadrupled?

motion remains S.H.M with time period = 2T

motion remains simple harmonic with time period = T/2

motion remains S.H.M with time period = 4T

motion remains S.H.M and period remains nearly constant

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Magnetism and Matter question

2003 · Q138

Magnetism and Matter question

2003 · Q138

View written solutionFree

Correct answer: B

The time period of a bar magnet in a magnetic field is given by

T = $2\pi \sqrt {{I \over {MB}}} = 2\pi \sqrt {{{m{k^2}} \over {MB}}} \propto \sqrt m$

$\therefore$ ${{{T_1}} \over {{T_2}}} = {{\sqrt {{m_1}} } \over {\sqrt {{m_2}} }}$ = ${{{T_1}} \over {{T_2}}} = {{\sqrt {{m_1}} } \over {\sqrt {{m_2}} }} = {{\sqrt m } \over {\sqrt {4m} }} = {1 \over 2}$

$\Rightarrow$ T₂ = 2T₁ = 2T.

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Sign in

Magnetism and Matter question

2003 · Q138

Magnetism and Matter question

2003 · Q138

View written solutionFree

Correct answer: B

The time period of a bar magnet in a magnetic field is given by

T = $2\pi \sqrt {{I \over {MB}}} = 2\pi \sqrt {{{m{k^2}} \over {MB}}} \propto \sqrt m$

$\therefore$ ${{{T_1}} \over {{T_2}}} = {{\sqrt {{m_1}} } \over {\sqrt {{m_2}} }}$ = ${{{T_1}} \over {{T_2}}} = {{\sqrt {{m_1}} } \over {\sqrt {{m_2}} }} = {{\sqrt m } \over {\sqrt {4m} }} = {1 \over 2}$

$\Rightarrow$ T₂ = 2T₁ = 2T.