A magnetic needle has magnetic moment 6.7×10⁻² A m² and moment of inertia 7.5×10⁻⁶ kg m². Its period of oscillation is 0.67 s. Then the corresponding magnetic field is _.

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(1) 0.01 T. Given: m=6.7×10⁻² A m², I=7.5×10⁻⁶ kg m², T=0.67 s. Idea: a magnetic needle turned a little way out of line and released swings back with period T=2π√I/(mB) - the angular version of a mass on a spring, with mB as the stiffness. Rearranged: B=(4π²I)/(mT²). Numerator: 4π²I=39.5×7.5×10⁻⁶=2.96×10⁻⁴. Denominator: mT²=6.7×10⁻²×(0.67)²=3.01×10⁻². B=(2.96×10⁻⁴)/(3.01×10⁻²)=9.8×10⁻³ T, which is 0.01 T to two figures.

A magnetic needle has magnetic moment 6.7×10⁻² A m² and moment of inertia 7.5×10⁻⁶ kg m². Its period of oscillation is 0.67 s. Then the corresponding magnetic field is _______.

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