A circular coil carrying a current of 2.5 A | vedclass

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(A) The time period of oscillation of a current-carrying coil in an external magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is th

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$\frac{T}{2}$

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(A) The time period of oscillation of a current-carrying coil in an external magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia,$M$ is the magnetic dipole moment,and $B$ is the magnetic field.Since $M = NIA$ (where $N$ is the number of turns,$I_{curr}$ is the current,and $A$ is the area),we have $M \propto I_{curr}$.Therefore,$T \propto \frac{1}{\sqrt{M}} \propto \frac{1}{\sqrt{I_{curr}}}$.Given $I_{curr1} = 2.5 \ A$ and $I_{curr2} = 10 \ A$,we have $\frac{T_2}{T_1} = \sqrt{\frac{I_{curr1}}{I_{curr2}}} = \sqrt{\frac{2.5}{10}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.Thus,$T_2 = \frac{T_1}{2} = \frac{T}{2}$.

$A$ circular coil carrying a current of $2.5 \ A$ is free to rotate about an axis in its plane perpendicular to an external magnetic field. When the coil is made to oscillate,the time period of oscillation is $T$. If the current through the coil is $10 \ A$,the time period of oscillation is

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