Earth s Magnetic Field Adapted by MMWaite from Measurement of Earth's Magnetic Field [Horizontal Component] by Dr. Harold Skelton

Transcription

1 Adapted by MMWaite from Measurement of Earth's Magnetic Field [Horizontal Component] by Dr. Harold Skelton Object: The purpose of this lab is to determine the horizontal component of the Earth s Magnetic Field here in West Chester, PA. Reference: Halliday & Resnick, Ch Apparatus: 1. Neodymium Magnets (Pasco EM-8621) 2. Compass 3. Ruler Earth s Magnetic Field Part : The Earth has many magnetic elements and materials in it s core. The effect of these magnetic materials is that the Earth has an intrinsic magnetic field around it. Although the details of this magnetic field are very complex, the field can essentially be modeled as a big bar magnet within the Earth as shown above. Here in West Chester, the magnetic field comes up out of the Earth s surface and curves over the surface. Thus, there are two components, one perpendicular to the Earth s surface (called the VERTCAL component), and one parallel to the Earth s surface (called the HORZONTAL component.) We will be determining the horizontal component of the field using what we know about placing magnetic dipoles in existing fields, and with what we know about adding two vectors (B-field vectors) together. First, we need to recall what happens when we place a magnetic dipole (a small magnet with a N and a S pole will suffice) in an existing magnetic field (the Earth s field, BE). Recall that, like electric charges, magnetic poles are attracted to opposites (N to S, or S to N) and repelled from similar (N vs. N, or S vs. S). Thus, a magnetic dipole placed in an existing magnetic field will experience a torque. where µ τ = µ is the magnetic dipole moment of the magnet. n terms of the magnitude of this torque, we get: τ = µ sinθ = α Thus, our dipole (magnet) will angularly accelerate, or rotate, to align itself with the field it has been placed in. (The minus sign has been introduced because this is a restoring torque, just like the spring force for a mass on a spring is a restoring force.) For small angles, we can make the small angle approximation sinθ θ, and we get: µ θ = α mwaite@wcupa.edu 1 March 16, 2010

2 [Compare this to the spring force (-kx) acting on a mass attached to a spring: kx = ma ] Now we can use the fact that the angular acceleration is the second derivative of angular position and we get a differential equation. d 2 θ dt + µ θ = 0 2 No worries here! We ve seen this before, we just need a function that, when we take it s second derivative, we get the function back with a minus sign... sine will do. d 2 θ ( t) dt 2 ( ) = θ 0 sin( ωt) θ t ( ) dt dθ t = ωθ 0 cos( ωt) = ω 2 θ 0 sin( ωt) = ω 2 θ ( t) f we plug this into the differential equation, we get: d 2 θ dt + µ θ = 0 2 ω 2 θ 0 cos( ωt) + µ θ 0 sin( ωt) = 0 ω 2 = µ n this equation, ω is just the angular frequency. The dipole will oscillate back and forth. We have what is called a torsional pendulum here. t s very much like a pendulum, but instead of swinging back and forth and back and forth, it spins back and forth and back and forth. f you recall from pendulums, the period of oscillation is directly related to the angular frequency as follows: Together, we get: T = 1 f = 2π ω ω = 2π T ω 2 = 4π 2 T 2 ( ω 2 = ) µ = 4π 2 T 2 Here we see that, if we measure the period of oscillation, and if we know the moment of inertia of the rotating dipole, we can solve for the product µ fairly easily. Earth s Magnetic Field The rotating dipole will be our small neodymium magnets. The moment of inertia is given by: mwaite@wcupa.edu 2 March 16, 2010

3 = 1 12 ml mr2 where m is the mass of the dipole, L is the length and R is the radius. have measured these values and they values are given in the chart below. Measurement Value Uncertainty m 44.5 g ±0.1 g L 21 mm 1 mm Part : R 9.5 mm 0.5 mm OK, so we have an expression for the product of the dipole moment and the Earth s B-field. But this is two unknowns and only one equation. We need another equation with these two variables in it. We will again use our neodymium magnets, along with a compass needle. A compass needle is essentially a small magnetic dipole itself. f a compass is placed in the Earth s magnetic field, it s needle will align itself with the field and point towards magnetic north. But, if we place another magnet nearby, the compass needle will sense two fields, experience two torques, and align itself with the net field, which will be a vector superposition of the two. At some distance from the magnet, the magnetic field can be approximated as that of the axial field of a current loop: = µ 0 µ 2π z 3 where z is the distance away, and µ is the magnetic dipole moment of the magnet. Thus, this magnet will exert a torque on the compass needle, just like the Earth s magnetic field does! So, lay a meter stick down on the table and place the compass on the end centered over a number you know. Now rotate the entire meter stick around until the compass needle points perpendicular to the meter stick as shown. The compass reading should be 0. Now, if you place the neodymium magnet on the meter stick about 20 cm away, you will notice that the compass needle has been deflected as shown below: mwaite@wcupa.edu 3 March 16, 2010

4 θ Now there are two components of the magnetic field at the compass, thus two torques, and the needle points in the direction of the net field. Once the compass needle stops rotating, the two torques must balance: τ Earth = µ compass sinθ τ magnet = µ compass sin( 90 θ) = µ compass cosθ µ compass sinθ = µ compass cosθ sinθ = cosθ = tanθ Let s plug this in the equation for the magnetic field from the middle of the previous page: tanθ = µ 0 2π µ f we change z, we will see different angular deflections, θ. The angle, θ, is a variable, dependent on the value of z. Also, note that tan θ and 1/z 3 are linearly related (in the form y = mx + b ). Thus, if we plot tan θ on the y-axis, and 1/z 3 on the x-axis, the slope will be all that stuff in the parentheses, and it will allow us to solve for µ!! 1 z 3 NOW we have two equations and two unknowns! You can use these two equations to eliminate µ and solve for. Voila! µ = 4π 2 tanθ = µ 0 T 2 2π µ 1 z 3 mwaite@wcupa.edu 4 March 16, 2010

5 Procedure: Earth s Magnetic Field 1) Suspend the neodymium magnet from a long(ish) string such that it hangs freely. t should be slightly rotating (twisting) back and forth. Record the time it takes for 10 oscillations. Repeat this 10 times and fill in the chart below: Trial Oscillations Time (s) Period (s) ) Calculate the average period; the uncertainty in the period will be the standard deviation of this result. 3) From this information, determine the product µ and its uncertainty, δ ( µ ). 4) Now, place the compass and neodymium magnet on the meter stick as describe in Part above. Place the magnet at 20 cm from the compass. Record the angular position of the compass needle. Move the magnet 5 cm further away and repeat. Fill in the chart below. At this point in the semester, you should be very aware of how to determine experimental uncertainties. t is up to YOU to determine how to make your estimate of the uncertainty in the angle, δθ, DO NOT ASK YOUR NSTRUCTOR! He/she will not tell you!! z (cm) z (cm) (degrees) (degrees) mwaite@wcupa.edu 5 March 16, 2010

6 z (cm) z (cm) (degrees) (degrees) ) Make a plot on Excel (or a similar program) of tan θ vs. 1/z 3 (tan θ on the y-axis, and 1/z 3 on the x- axis.) 6) Determine the slope of the plot and use this result to find the ratio µ as described above. 7) Use the two results in steps (3) and (6) to calculate BE. 8) Calculate the uncertainty in BE. How does this compare to other values? (Go to Use ZP code if doesn t work.) To be turned in next week: Abstract Charts Excel Plot (or similar program) Error propagation calculations (doesn t need to be typed, but must be neat!) Comparison to the NOAA value, including a discussion of any discrepancies. mwaite@wcupa.edu 6 March 16, 2010