1. Basándose en la figura siguiente, si el radio de curvatura del alambre es de 50 cm, determine la magnitud de la campo magnético en el centro de curvatura (en el punto 0, véase la figura siguiente). (µo = 4π.10-7 Wb.A-1 m-1)
Conocido :
Radio (r) = 50 cm = 0.5 m
Corriente eléctrica (I) = 1.5 Amperios
La permeabilidad al vacío (μo) = 4π.10-7 Wb.A-1 m-1
Querido: Tla magnitud del campo magnético
solución:
360o = 1 circunferencia de un circulo. 120o / 360o = 1/3 entonces 120o = 1/3 x circunferencia of a circle.
The equation of the magnetic field at the center of the coil with several loops :
B = the magnitude of the magnetic field, N = number of loops, I = electric current, r = radius of curvature
In the above problem, there is only one loop so that N is eliminated from the equation. The wire coil on the above problem is not 1 circle but 1/3 circle :
The magnitude of the magnetic field at the center of curvature :
2. Based on the figure below, the electric current flows in the wire is 6-A and radius of curvature is R = 3π cm, to determine the magnitude of the magnetic field at point P.
Conocido :
Radius of curvature (r) = 3π cm = (3π/100) m
= 3π/102 m = 3π.10-2 m
Electric current (I) = 6 A
La permeabilidad al vacío (μo) = 4π.10-7 Wb.A-1 m-1
Buscado : The magnitude of the magnetic field
solución:
360o = 1 circunferencia of a circle. 45o / 360o = 1/8 entonces 45o = 1 / 8 x circunferencia of a circle.
The magnitude of the magnetic field at the center of curvature :
3. Electric current flows in wire = 9-A, the radius of curvature (R) = 2π cm and µo = 4π.10-7 Wb.A-1.m-1, determine the magnitude of the magnetic field at point P.
Conocido :
Radius of curvature (r) = 2π cm = (2π/100) m
= 2π/102 m = 2π.10-2 m
Electric current (I) = 9 A
La permeabilidad al vacío (μo) = 4π.10-7 Wb.A-1 m-1
Buscado : The magnitude of the magnetic field at point P
solución:
360o 120o = 240o. 240o / 360o = 2/3 entonces 240o = 2/3 x circunferencia of a circle.
The magnitude of the magnetic field at the center of curvature :
