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Questions 1. When you fit the 1 / T? vs Ic curves obtained from Measurements #1, #2, #3 on Page #1 you cannot take out Ic = 0 point. Explain why at I = 0 the iron rod still rotates? 3 . What are the SI units of the fitting slope and intercept in Fig. 4 and Table 1? (Show your work).4. What is the SI units of the constant C in Table 2? (Show your work). 5. Why does the bar magnetic rotate rather than oscillate at DC Volt = 1 V in Measurement #4?\fEarth's magnetic field Purpose Obtain the horizontal component of the earth's magnetic field ( B, ) by measuring the period of oscillation of a bar magnet in a known field. Equipment PASCO Helmholtz coils (EM-671 1and EM-6715 N=200, radius R=10.5 cm), bar magnet, photogate (ME- 9204A), resistor ~ 40 $2 (2W), PASCO 850 interface, computer. Theory Except at the magnetic equator, the earth's magnetic field is not horizontal. In South Florida, the magnitude of the earth's field is about 0.5 Gauss while horizontal component is about 0.25 gauss. The horizontal component of the earth's magnetic field ( B, ) can be obtained by studying how it affects a bar magnet which is suspended so that it is free to rotate in a horizontal plane. Such a magnet will line up to in the direction of the horizontal component of the earth's field. That is what a compass is! The reason for this is that a magnetic field of magnitude B will produce a torque t on a bar magnet given by: T = -mBsino, (1 ) where the minus sign means that it is a restoring torque, m is the magnitude of the magnetic moment of the bar magnet, 0 is the angle between the magnetic moment and magnetic field vector, as shown in Fig. 1. m bar magnet B Figure 1 A bar magnet in an external B -field If the angle O is small, then sin0 20 and equation (1) reduces to T =-mBe (2) This form is similar to F = -k x in the case of a simple harmonic oscillator such as a mass hanging from a spring or a simple pendulum. If we set the bar magnet into oscillation, the motion of the bar magnet is approximately harmonic and the equation of motion is given by the Newton's second law for a rigid body t = la = 1- (3) dt 2 d'0 where I is the moment of inertia and a dt 2 ", the angular acceleration of the bar magnet. Substituting (2) into (3): -mB0 =I- d'0 mB or dt2 dt 2 +-0=0, or d'0 + w 0 = 0 mB with dt 2 I (4 ) where @ is the angular frequency of the oscillation.2TT W m - B = CB, m T2 472 I with C = (5) T is the period of the oscillation. The constant C depends on the magnetic moment and the moment of inertia of the bar magnet, which is unknown. Hence, in order to obtain the horizontal component of the earth's field, we must first find C for the bar magnet, i.e., we must calibrate it in a known uniform magnetic field. In this experiment the known uniform magnetic field is produced by Helmholtz coils consisting two identical coils separated by a distance equal to their radius. The two coils carry the same current in a direction such that, at the midpoint between them on their axis, their magnetic field will have the same direction and the same magnitude, so the magnitude of the total magnetic field at that point is B = 2 N Molc - 0.72N Hold (6) R R N is the total # of turns of each Helmholtz coil, R is the radius of Helmholtz coil, I is the DC current going through each Helmholtz coil, / = 4x x10" T. m/A. The SI units (international system of units) of the magnetic field is T (Tesla). The Gaussian units of the magnetic field is G (Gauss). 1 T = 10000 G. If we align the Helmholtz coils so that its axis is in the same direction as B, , the horizontal component 0.72N Hold of the earth's magnetic field, the fields add to give a total magnetic field Brot = By + and R equation (5) becomes T2 = CBlot = C Bn + 0.7 0.72 NHOLC = CB, +CU.INHo Ic (7 ) R This equation shows how we can find the constant C and determine the magnetic field B, . We cannot obtain C from a single measurement because both C and B, are unknown. We must measure the period of oscillation T for various values of the current I through the coils. The plot of versus Ic will be a straight line with a slope CO.12N Loan and an intercept CB, from which the values of C and B, can be R determined since the Helmholtz coils used in this experiment with /V=200 turns, R =10.5 cm