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A student performs a lab using a pair of coils with radius R = 10.5 cm and N = 95 windings. Using the equation in the theory, calculate the magnetic field at the center of both coils when the coil current is 4.7 A

A student performs a lab using a pair of coils with radius R = 10.5 cm and N = 95 windings. Using the equation in the theory, calculate the magnetic field at the center of both coils when the coil current is 4.7 A

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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A student performs a lab using a pair of coils with radius R = 10.5 cm and N = 95 windings. Using the equation in the theory, calculate the magnetic field at the center of both coils when the coil current is 4.7 A.

THEORY When an electron is accelerated through a potential (voltage) difference V, the work done on the electron (eV) equals the change in kinetic energy 2eV 1 -mv² = eV ⇒V= 2 (1) where m is the electron mass, e is the elementary charge and v the electron velocity. An electron moving perpendicular to a magnetic field B is subject to a force of magnitude 2eV F = eBv = eB (2) m which is perpendicular to the electron velocity and the magnetic field, and therefore causes the electron to move in a circle. The centripetal force required for a circular orbit of radius r is F=m- (3) r Equating the two expressions for F and solving for velocity gives V² eBr F=m r m Combining the above with equation 1 leads to an equation for the charge-to-mass ratio 2eV eBr e 2V B(z) m = = = (5) m m m B²r² If the voltage (V), the magnetic field (B) and the radius of the circular electron path (r) are measured, the charge-to-mass ratio can be calculated using this equation. eBv ⇒v= The magnetic field that the electron moves through is created using two parallel coils (see the photos on the next page). At a distance z from the center of a single coil the field is HoNIR ² (6) B = 2- 2[R² +2²7³/² where μ. = 4 x 10-7 T-m/A, R is the radius of the coil, N is the number of times the wire is wound around the coil, and I is the current. In this case we have two coils, both a distance z = R/2 from the electron beam in the vacuum tube. The field is thus HoNIR 2 R Por (9)7² R² + 2 (4) ⇒ B= 8 MONI 125 R (7)
Transcribed Image Text:THEORY When an electron is accelerated through a potential (voltage) difference V, the work done on the electron (eV) equals the change in kinetic energy 2eV 1 -mv² = eV ⇒V= 2 (1) where m is the electron mass, e is the elementary charge and v the electron velocity. An electron moving perpendicular to a magnetic field B is subject to a force of magnitude 2eV F = eBv = eB (2) m which is perpendicular to the electron velocity and the magnetic field, and therefore causes the electron to move in a circle. The centripetal force required for a circular orbit of radius r is F=m- (3) r Equating the two expressions for F and solving for velocity gives V² eBr F=m r m Combining the above with equation 1 leads to an equation for the charge-to-mass ratio 2eV eBr e 2V B(z) m = = = (5) m m m B²r² If the voltage (V), the magnetic field (B) and the radius of the circular electron path (r) are measured, the charge-to-mass ratio can be calculated using this equation. eBv ⇒v= The magnetic field that the electron moves through is created using two parallel coils (see the photos on the next page). At a distance z from the center of a single coil the field is HoNIR ² (6) B = 2- 2[R² +2²7³/² where μ. = 4 x 10-7 T-m/A, R is the radius of the coil, N is the number of times the wire is wound around the coil, and I is the current. In this case we have two coils, both a distance z = R/2 from the electron beam in the vacuum tube. The field is thus HoNIR 2 R Por (9)7² R² + 2 (4) ⇒ B= 8 MONI 125 R (7)
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