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www.GOALias.blogspot.comPhysicsIn magnitude τ = mB sinθHere τ is restoring torque and θ is the angle between m and B.2d θTherefore, in equilibrium I =−mBsinθ2dtNegative sign with mB sinθ implies that restoring torque is in oppositionto deflecting torque. For small values of θ in radians, we approximatesin θ ≈ θ and get2d θI ≈ – mB θ2dt2d θ mBor, =− θ2dt IThis represents a simple harmonic motion. The square of the angularfrequency is ω 2 = mB/I and the time period is,IT = 2πmB4 π2 I2(5.4)or B =(5.5)mTAn expression for magnetic potential energy can also be obtained onlines similar to electrostatic potential energy.The magnetic potential energy U mis given byUm= ∫ τθ ( ) d θ= ∫ mB sin θ =−mB cosθ=−mB g (5.6)We have emphasised in Chapter 2 that the zero of potential energycan be fixed at one’s convenience. Taking the constant of integration to bezero means fixing the zero of potential energy at θ = 90º, i.e., when theneedle is perpendicular to the field. Equation (5.6) shows that potentialenergy is minimum (= –mB) at θ = 0º (most stable position) and maximum(= +mB) at θ = 180º (most unstable position).Example 5.1 In Fig. 5.4(b), the magnetic needle has magnetic moment6.7 × 10 –2 Am 2 and moment of inertia I = 7.5 × 10 –6 kg m 2 . It performs10 complete oscillations in 6.70 s. What is the magnitude of themagnetic field?178EXAMPLE 5.1Solution The time period of oscillation is,6.70T = = 0.67s10From Eq. (5.5)24πIB =2mT=4 × (3.14) × 7.5 × 10–2 26.7 × 10 × (0.67)= 0.01 T2 −6