Question

An alpha particle (consisting of two protons and two neutrons) is moving in a circle at a constant speed, perpendicular to a uniform magnetic field applied by some current-carrying coils. The alpha particle makes one clockwise revolution every 81 nanoseconds.
If the speed is small compared to the speed of light, what is the numerical magnitude B of the magnetic field made by the coils? What is the direction of this magnetic field?

Answer 1

Answer:

1.64 T direction of the magnetic field is into the page .

Explanation:

Solution  

The magnetic force F_B causes the a-particle to move in a circular motion where the a-particle gains a centrifugal force F_C So, the magnetic force and the centrifugal force are equal to each other.  

F_B = F_C                                                   (1)

The magnetic force is affected by the magnitude and the direction of the magnetic force, the direction of the magnetic field, the sign of the charge and the direction of the moving of the charge and it is given by  

F_B = qvB

The centrifugal force is related to the mass of the a-particle ma and the circular radius R by  

F_C = m_a*v^2/R

Now let us plug the expressions of F_C and F_B into equation (1) to get the magnetic field B  

F_B = F_C      

qvB =m_a*v^2/R                                         (solve for B)

B = m_a*v^2/qR                                           (2)

The term {v/R) equals the angular frequency which equals 2$\pi$/T . From the next steps, we got this relationship  

v = x/T = 2$\pi$R/T

v/R = 2$\pi$/T

Where the distance x equals the circumference of the circle where the alpha particle moves. Hence, the magnetic field is given by

B = (m_a/q)*(v/R)

  = (m_a/q)*2$\pi$/T                                           (3)

Where T is the time, mo, is the mass of the alpha particle and q is the charge of the alpha particle and equals 2e.  

Alpha particle consists of two protons and two neutrons, therefore, the total mass of the alpha particle is given by  

m_a = 2m_p+ 2m_n

       = 2 (1.672 x 10^-27 kg) +2 (1.675 x 10^-27 kg)

       = 6.7 x 10^-27 kg  

Now we can plug our values for m_a,q and T into equation (3) to get the magnetic field B  

B = m_a*2$\pi$/q*T

   = 2$\pi$(6.7*10^-27)/2(1.6*10^-19)(81*10^-9s)

   = 1.64 T

The right-hand rule determines the direction of the magnetic field B and the direction of the motion of the alpha particle. Where your thump is in the direction of the magnetic field, while your remaining fingers curl in the direction of the motion of the alpha particle (Clockwise). Let us apply this rule, we find that the direction of the magnetic field is into the page .  

Answer 2

Answer:

B = 4.41 10⁻⁴ T

The direction of the magnetic field is perpendicular to the speed in such a way that if the force is in the clokwise direction

Explanation:

For this problem let's use Newton's second law where the force is magnetic

        F = ma

the magnetic force is

        F = q v x B

The bold are vectors, as they indicate that the magnetic field and velocity are perpendicular the magnitude of the force is

        F = q v B

acceleration is centripetal

        a = v² / r

we substitute

        q v B = m v² / r

         B = m v / q r

Since the velocity modulus is constant, we can use the relation of uniform motion with the time of a lap that we call period (T)

       v = d / t

   

the length of the circle is

       d = 2π r

       v = 2π r / T

we substitute in the magnetic field equation

       B = m /qr  (2π r / T)

       B = m/q  2π / T

we calculate

       B = 2  9.1 10⁻³¹ / 2 1.6 10⁻¹⁹    2π / 81 10⁻⁹

       B = 4.41 10⁻⁴ T

     

The direction of the magnetic field is perpendicular to the speed in such a way that if the force is in the clokwise direction, for example if the speed goes on the positive axis the field goes up