Question

A charged particle moving through a magnetic field at night angles to the field with a speed of 24.7 m/s experiences a magnetic force of 2.78 x 10 N. Determine the magnetic force on an identical particle when it travels through the same magnetic field with a speed of 5.44 m/s at an angle of 31.20 relative to the magnetic field. x How does the magnetic force acting on a charged particle moving through a magnetic field depend on the charge of the particle, speed of the particle, magnitude of the magnetic field, and the direction the particle is traveling through the magnetic field? For this problem, why have we not been given values for the charge on the particles and the magnetic field? N Additional Materials cBook Tutorial E Show My Work (Optionalt Submit Answer 1/1 Points) DETAILS PREVIOUS ANSWERS OSUNIPHYS1 28.2.WA.007. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER Using appropriate technology, you determine that the Earth's magnetic field at your location has a magnitude of 5.90 x 10-5T and is directed north at an angle of 62° below the horizontal. You charge a metallic projectile and fire it north with a speed of 640 m/s and at an angle of 14° above the horizontal. Determine the sign (+ or -) and charge of the projectile, if the magnetic force on the projectile is 2.50 x 10 10 N, directed due west. x How is the magnetic force on a moving charged object related to the magnitude of the charge on the object, the speed of the object, the magnitude of the magnetic field experienced by the object, and the angle between the direction of the velocity of the object and the magnetic field? What does RHR tell us about the direction of the magnetic force acting on a charged object moving through a magnetic field? Additional Materials

Answer 1

We need to apply the formula only. The relation between magnetic field, magnetic force, charge on the particle and the velocity of the particle is given at the very beginning of the solution. For this problem, we do not need directly the charge and the magnetic field values, that is because, we have the value of their multiplication, and that value we can use directly for solving the question. No need to have seperate values of charge and magnetic field.

1. 
   B = >7= 24.7 m/s Magnetic force q (DXB) Fm =q VB sine Here, e=90 sino = 1. Fm = q VB. 2.78X DOWN = 9B (24.7) qB Ns/m 24.7 9B 1. 12 x 10 Ns/m = 2.78x104 -0 Now, when 0 = 31.2° R. = 1544 m/s 131.2 Em = q VB sine = (9B) (5.44 mb) six 3124 +9B) (5.44 m/s) (0.52) putting the value of GB) from eq" . Em = (1.12 X10% Ns/m) (5.44 m/s) (0.52) Im = 3.17 x 10 N