A corner point makes angles θ_{1} and θ_{2} with the ends of a straight wire carrying current I. The distance of the point from the wire is a. What is the

magnetic field at the point?

# magnetic field

# Radius of the circular path described by a proton

# Particle enters a magnetic field. Question 3.

An electron is accelerated from rest through a potential difference of 200 V and then enters a magnetic field of strength 0.02 T acting at right angles to its path. Calculate the radius of the resulting electron orbit.

Solution.

Based on the Law of Conservation of Energy

Ep – Electric potential energy ; Ek – Mechanical Kinetic energy

Ep = Ek

Ep = q*U ; U=200 v

Ek = m * v^2 / 2

q*U = m * v^2 / 2

Now we could find a velocity. v = SQRT(2*q*U/m)

Lorentz force. It is a magnetic force on a point charge.

F = q * v* B * sin (A)

F – centripetal force equal :

F= ma = m * v^2/r = q * v* B * sin (A), sin (90) = 1

Now, we could find another equation for the velocity

v = q*B*r/m

Finally, we have everything in order to calculate radius.

We also could use this equations and get another equation.

v=SQRT(2*q*U/m)

v = q*B*r/m

So, r= SQRT(2*U*m/(q*B^2) ) = SQRT (2*200*9.1*10^-31/(1.6*10^-19*0.02^2)) = 0.002 m

# Particle enters a magnetic field. Question 2.

Question – Source: Yahoo! Answers !

A proton, mass 1.67 · 10 -27 kg and charge +1.6 · 10 -19 C, moves in a circular orbit perpendicular to a uniform magnetic field of 0.82 T.

Find the time for the proton to make one complete circular orbit.

Solution.

The Newton’s Second Law : F=m*a

Circular motion formulas:

a= v^2/r ; v=2*pi*r/T

Electromagnetic force on the proton (Lorentz Force) :

F=q*v*B*sin(a);

sin(a) = 1 because circular orbit perpendicular to a uniform magnetic field ;

So,

q*v*B=m*v^2/r

q*B=m*v/r

q*B= m*2*pi*r/(r*T)

T=2*pi*m/(q*B) ; the time for the proton to make one complete circular orbit;

T= 2*3.14*1.67*10^-27/(1.6*10^-19*0.82) = 8*10^-8 s

# Particle enters a magnetic field. Question 1.

Question. Source. Yahoo ! Answers !

A singly charged positive ion has a mass of 2.64 10-26 kg. After being accelerated through a potential difference of 240 V, the ion enters a magnetic field of 0.530 T, in a direction perpendicular to the field. Calculate the radius of the path of the ion in the field.

Solution.

You have to use the Law of Conservation of Energy

Ep – Electric potential energy ; Ek – Mechanical Kinetic energy

Ep = Ek

Ep = q*U

Ek = m * v^2 / 2

q*U = m * v^2 / 2

So, you could find that velocity. v = SQRT(2*q*U/m)

F – centripetal force equal :

F= ma = m * v^2/r = q * v* B * sin (A), sin (90) = 1

So, you could find another equation for the velocity

v = q*B*r/m

Now, You have everything in order to calculate radius.

You also could use this equations and get another equation.

v=SQRT(2*q*U/m)

v = q*B*r/m

So, r= SQRT(2*U*m/(q*B^2)