The magnetic needle has magnetic moment 6.7×10−2Am2 and Moment of Inertia 7.5×10−6kgm2 , if the time period of oscillation is 0.67 sec then the magnetic field ‘B’ is

T=2πIMB on sub B=4π2IMT2=4X3,142X7.5X10−66.7X10−2X0.672=0.01T

The magnetic needle has magnetic moment 6.7×10−2Am2 and Moment of Inertia 7.5×10−6kgm2 , if the time period of oscillation is 0.67 sec then the magnetic field ‘B’ is

T=2πIMB on sub B=4π2IMT2=4X3,142X7.5X10−66.7X10−2X0.672=0.01T

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The magnetic needle has magnetic moment $6.7 \times 10^{- 2} {Am}^{2}$ and Moment of Inertia $7.5 \times 10^{- 6} {kgm}^{2}$ , if the time period of oscillation is 0.67 sec then the magnetic field ‘B’ is

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0.1T

0.02T

0.01T

0.2T

answer is D.

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$\begin{array}{l} T = 2 π \sqrt{\frac{I}{M B}} o n s u b B = \frac{4 π^{2} I}{M T^{2}} \\ = \frac{4 X {(3, 14)}^{2} X 7.5 X 10^{- 6}}{6.7 X 10^{- 2} X {(0.67)}^{2}} = 0.01 T \end{array}$

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The magnetic needle has magnetic moment 6.7×10−2Am2 and Moment of Inertia 7.5×10−6kgm2 , if the time period of oscillation is 0.67 sec then the magnetic field ‘B’ is

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The magnetic needle has magnetic moment $6.7 \times 10^{- 2} {Am}^{2}$ and Moment of Inertia $7.5 \times 10^{- 6} {kgm}^{2}$ , if the time period of oscillation is 0.67 sec then the magnetic field ‘B’ is