Wize University Physics Textbook (Master) > Newton's Laws of Motion: Forces and Dynamics

Magnetic Force

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Magnetic Force

An electric charge moving in a magnetic field experiences a force whose direction is perpendicular to its velocity and magnetic field direction.

F⃗=qV⃗×B⃗=qVBsin⁡(θ)\vec{F}=q\vec{V}\times\vec{B}=qVB\sin(\theta)F=qV×B=qVBsin(θ)
               Where θ\thetaθ is  angle between V and B  

Direction of force can be found by using right hand rule.
Right Hand Rule is designed for positive charges. If charge is negative, direction of the force will be opposite of what right hand rule gives.

Right Hand Rule for Magnetic Force:
 Notes
If v and B are parallel, then force is ZERO.
If v and B are perpendicular, then F = qvB.
Uncharged particles do not deviate in magnetic field  

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The charged particles are deflected in a curve when entering a region of constant magnetic field.

Acceleration in magnetic field
Acceleration due to magnetic force is perpendicular to the direction of the velocity. Therefore, it changes the direction of the velocity vector however the speed remains constant.

The charge particle will undergo uniform circular motion in a uniform magnetic field.

The centripetal acceleration is: ac = v2/R

The radius of rotation is: R = mv/qB

If there is both electric field and magnetic fields present in space, the force on the particle could be found using Lorentz force: 

F⃗=qE⃗+qv⃗×B⃗\vec{F}=q\vec{E}+q\vec{v}\times \vec{B}F=qE+qv×B

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An electron moves perpendicular to a uniform magnetic field which exerts a magnetic force on the electron shown in the figures below. In each case, find the direction of the magnetic field? 

 For the diagram on left side: 
 Right hand rule (RHR) for positive charge is  F→=qv→×B→\overrightarrow{F}=q\overrightarrow{v}\times\overrightarrow{B}F=qv×B
 If the charge is negative, the final answer should be flipped. 
 Thus, if this was a positive charge, the F would have been along +Y axis. This mean, RHR indicated that B must be into the page or -X axis. 
 

For the diagram on right side:
 Right hand rule (RHR) for positive charge is F→=qv→×B→\overrightarrow{F}=q\overrightarrow{v}\times\overrightarrow{B}F=qv×B 
 If the charge is negative, the final answer should be flipped. 
 Thus, if this was a positive charge, the F would have been along +Y axis. This mean, RHR indicated that B must be into the page or -Z axis. 

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a) I figure (a), an electron is moving with velocityvvvalong +x-axis through a uniform electric fieldEEE(along +y-axis). Find the direction of the magnetic fieldBBB

b) In figure (b), the magnitudes and directions ofBBBand EEEare given. Find the speedvvvsuch that the electron remains traveling along the x direction. 

Solution:

a) F⃗E\vec{F}_EFE​ is downward
F⃗B\vec{F}_BFB​ should be upward to haveFnet⃗=0→B\vec{F_{net}}=0\to BFnet​​=0→B is outward!

b)


   FE=FB=∣evB∣F_E=F_B=\left|evB\right|FE​=FB​=∣evB∣ 
  →Ee=evB→V =EB=90×1030.5=180×103 ms\begin{array}{ll}
\to Ee=evB \\
\to V\ =\frac{E}{B}&=\frac{90\times10^3}{0.5}\\
&=180\times10^3\ \frac{m}{s}
\end{array}→Ee=evB→V =BE​​=0.590×103​=180×103 sm​​

In figure (a), an electron is moving with velocity along +x-axis through uniform electric and magnetic fields. Find the direction of the magnetic field.
In figure (b), the magnitudes and directions of B and  EB\ and\ \ EB and  Eare given. Find the speed vvv such that the electron remains traveling along the x direction.
Now imagine the electric field is turned off. Suppose the electron moves at a speed of v=2.5×106 [m/s]v=2.5\times10^6\ \left[m/s\right]v=2.5×106 [m/s]and the magnetic field is B=1.4×10−4 [T]B=1.4\times10^{-4}\ \left[T\right]B=1.4×10−4 [T]. What is the radius of the circular path of the electron?
 What happens if there is a proton instead of electron in figure (b)? (Qualitative answer only, no calculations.)

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Practice: Magnetic Field of Current-Carrying Wires

Four straight wires, sitting at each corner of a square, are carrying equal electric currents I perpendicular to the page, as shown below. 

a) Find the direction of net magnetic field at the center of square. 
b) What is the expression for the magnitude of magnetic field at the center of the square?

A positively charged particle of 2.01 x 10-5 C and mass 7.52 x 10-16 kg is travelling at a velocity of 2.0 x 102 m/s in the positive horizontal direction. It enters a uniform magnetic field of 1.50 T which is pointing upward. 
How much is the force on the particle as it enters the magnetic field?
The particle starts to move in a circular path. Find the radius of this circular path.

FB=2.22⋅10−3 N , R = 17nmF_B=2.22\cdot10^{-3}\ N\ ,\ R\ =\ 17nmFB​=2.22⋅10−3 N , R = 17nm

FB=4⋅10−2 N , R = 22nmF_B=4\cdot10^{-2}\ N\ ,\ R\ =\ 22nmFB​=4⋅10−2 N , R = 22nm

FB=6.03⋅10−3 N , R = 5nmF_B=6.03\cdot10^{-3}\ N\ ,\ R\ =\ 5nmFB​=6.03⋅10−3 N , R = 5nm

FB=5.4⋅10−3 N , R = 12nmF_B=5.4\cdot10^{-3}\ N\ ,\ R\ =\ 12nmFB​=5.4⋅10−3 N , R = 12nm

I don't know

Extra Practice

EXAMPLE:  A positively charged particle of 2.01 x 10-5 C and mass 7.52 x 10-16 kg is travelling at a velocity of 2.0 x 102 m/s in the positive horizontal direction. It enters a uniform magnetic field of 1.50 T which is pointing upward. 
How much is the force on the particle as it enters the magnetic field?
The particle starts to move in a circular path. Find the radius of this circular path.

A positively charged particle of 2.01 x 10-5 C and mass 7.52 x 10-16 kg is travelling at a velocity of 2.0 x 102 m/s in the positive horizontal direction. It enters a uniform magnetic field of 1.50 T which is pointing upward. 
How much is the force on the particle as it enters the magnetic field?
The particle starts to move in a circular path. Find the radius of this circular path.

Practice: Magnetic Field of Current-Carrying Wires
Four straight wires, sitting at each corner of a square, are carrying equal electric currents I perpendicular to the page, as shown below. 
a) Find the direction of net magnetic field at the center of square. 

Practice: Magnetic Field of Current-Carrying Wires
Four straight wires, sitting at each corner of a square, are carrying equal electric currents I perpendicular to the page, as shown below. 
a) Find the direction of net magnetic field at the center of square. 

A positively charged particle of 2.01 x 10-5 C and mass 7.52 x 10-16 kg is travelling at a velocity of 2.0 x 102 m/s in the positive horizontal direction. It enters a uniform magnetic field of 1.50 T which is pointing upward. 
How much is the force on the particle as it enters the magnetic field?
The particle starts to move in a circular path. Find the radius of this circular path.

An ion with charge of 3.2 x 10-19 C and mass of 6.64 x 10-27 kg starts in a region with E=500 N/C k^\hat kk^ .  It moves for Δz=3m k^\Delta z=3m\,\hat kΔz=3mk^  and then experiences a constant magnetic field.
With what speed does it enter the 	magnetic field region?
What magnetic field is required so that the ion continues with a constant velocity?  

 An ion with charge of 3.2×10−19 C3.2\times 10^{-19}~\rm C3.2×10−19 C  and mass of  6.64×10−27 kg6.64\times 10^{-27}~\rm kg6.64×10−27 kg  is moving with velocity of 8×104 m/s ȷ^8\times 10^4~\rm m/s~\hat\jmath8×104 m/s ^​  in a region where there is a uniform magnetic field with intensity of −1 T ȷ^-1~\rm T~\hat\jmath−1 T ^​ 
What is the radius of rotation of the ion?
What is the vector that connects the ion to the center of the circle?

Magnetic Field of Current Carrying Wires

Four straight wires, sitting at corners of a square, are carrying equal electric currents perpendicular to this page, as shown below.  
Find the direction of net magnetic field at the center of square. 
What is the expression for the magnitude of magnetic field at the center? 