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5 Common Problems of Magnetic Force

physics.fisikastudycenter.com- 5 common problems of lorentz force that is acting in a moving particle with electric charge in it. Magnetic fields do involve in here.



Question 1
An electron with 1,6 x 10−19 C of charge moves with speed of 5 x 105 m/s into magnetic field of 0,8 T as shown below.


Find :
a) the magnitude of lorentz force on the electron
b) direction of magnetic force

Discussion
a) the magnitude of lorentz force on the electron
By using the magnetic force formula
F = BQV sin θ
where
B -magnetic field (Tesla), Q -particle charge (Coulomb), ν particle velocity (m/s) and θ -the angle between velocity and magnetic field direction. In this case, θ = 90°.
F = (0,8)(1,6 x 10−19)(5 x 105)(1) = 6,4 x 10−14 Newton

b) direction of magnetic force
Use the right hand rules for magnetic force:


4 fingers = magnetic field direction
the thumb = velocity
palm = magnetic force dirwection → for positive (+) charges, take the opposite direction for the negative ones.
Take a note U is the north pole of magnet, and S is the south pole of magnet so then we have the magnetic field comes from U to the S. Apply the right hand rules and we have the direction of magnetic force that is acting on the electron (negative charge) is toward the reader.

Question 2
A positron of 1,6 x 10−19 C travels with velocity of 5 x 105 m/s into 0,8 T of magnetic field s shown in our figure below.


Find :
a) the magnetic force
b) the direction of magnetic force

Discussion
a) F = (0,8)(1,6 x 10−19)(5 x 105)(1) = 6,4 x 10−14 Newton
b) Positron is a positive charge, so by applying the right hand rules we get the direction of magnetic force is into reading sheet, leaving the reader.




Problem 3
A conductor of 15 A of current and its direction shown below. A positive charge of 0.4 C that is 8 mm away from the conductor travels with its velocity of 5 x 103 m/s.



Find the magnetic force acting on this charge!

Pembahasan
Firtsly find the magnetic field producd by the conductor at distance of 8 mm, and then use magnetic formula to get the result:

B = μoI/2πa

B = (4π x 10−7)(15)/(2π)(8 x 10−3)
B = (15/4) x 10−4 Tesla
F = BQV sin 90°
F = ((15/4) x 10−4 )(0,4)(5 x 103)(1) = 0,75 Newton


Problem 4
Two charges that are Q1 = 2Q and Q2 = Q with the masses of m1 = m and m2 = 2 m travel in the same velocity into a homogenous magnetic field B. Compare the radius of trajectory that are formed by the two charges!

Solution
Consider this case as circular motion involve the magnetic force:



Question 5
A particle with electric charge of Q travels into a magnetic field B and electric field E with velocity of 2 x 104 m/s . The presence of magnetic field and electric field doesn't influence complitely to the particle motion. Compare the magnitude of magnetic and electric fields in this case!

Solution
The magnetic force and the electric force that are acting on the particle must be in the same magnitude and at the opposite of directions so the net resultant of force is zero.

BQV = QE
B/E = 1 / (2 x 104)
B/E = 0,5 x 10− 4 TC/N

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