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In this Physics tutorial, you will learn:

- How to find the magnetic force acting a moving charge inside a magnetic field?
- The same for the direction of this magnetic force
- How does a charged particle move when it is inside a magnetic field?
- How to find the period and radius of trajectory of a charged object inside a magnetic field?
- What is Lorentz Force? How to calculate it?

F_{1} = F_tot/n

Since the above magnetic force is the same magnetic force we have discussed in the previous tutorial (Ampere's Force), we can write F_{1} = (I ∙ B ∙ L)/n

Giving that the current I is I = ∆Q/∆t

where ΔQ is the charge flowing through the wire in the time interval Δt and L = v ∙ ∆t

where v is the velocity of moving charges throughout the wire, we obtain F_{1} = (∆Q/∆t ∙ B ∙ v ∙ ∆t)/n = (∆Q ∙ B ∙ v)/n

In addition, the total charge flowing through the wire during the interval Δt is a multiple of elementary charge e (i.e. ΔQ = n ∙ e) we obtain F_{1} = (n ∙ e ∙ B ∙ v)/n = e ∙ B ∙ v

The vector form of the above equation is F*⃗*_{1} = e ∙ (v*⃗* × B*⃗*)

Remark! The symbol e used above is not intended for electrons but for elementary charges despite in general there are the electrons the changes that can move. We have always taken the direction of current from positive to negative. Therefore, to find the direction of magnetic force acting on a moving elementary charge we must assume it as positive, although we know that only negative charges (electrons) are able to move through a conductor (positive charges can move only inside an electrolyte). Since the above formula derives from that of Ampere's Force, the magnetic force on positive charges is found by using the Fleming's left hand rule. According this rule the four fingers lie in the direction of motion of the positive charges, the palm is punched by magnetic field lines and the thumb shows the moving direction caused on the wire due to this interaction. But if the type of charge changes (i.e. if we consider a negative charge instead of a positive one), the direction of force will change as well. For example, the wire in the figure shown earlier moves in the onto-the-page direction. This is illustrated in the figure below. Not always the direction of particles motion is perpendicular to the magnetic field lines. When these two vectors form another angle θ to each other, we have to consider this angle as well. The formula of magnetic force for an elementary electric charge in such conditions therefore becomes F_{1} = e ∙ v ∙ B ∙ sinθ

However, the force vector will still be perpendicular to the plane of the other two vectors (v and B). Example: An electron is moving at 200 m/s at 300 to the direction of a 6mT magnetic field lines as shown in the figure. a) What is the magnitude of magnetic force produced? b) What is the direction of this magnetic force? Take the magnitude of elementary charge equal to 1.6 × 10-19 C. Solution a) Using the scalar equation for the force acting on an elementary charge when moving inside a magnetic field F_{1} = e ∙ v ∙ B ∙ sinθ

we obtain after substitutions F_{1} = (1.6 × 10^(-19) C) ∙ (2 × 10^{2} m/s) ∙ (6 × 10^{-3} T) ∙ sinθ = 1.92 × 10^(-19) N

b) The direction of magnetic force is found by using the Fleming's Left Hand Rule for a positive charge and then taking the opposite direction of it. We have to consider the vertical component of velocity for orientation. Here the direction of velocity replaces the direction of electric current in Ampere's Force (this is obvious since current represents a movement of electric charges). Thus, the four fingers are directed downwards while the palm is directed due right as it is punched by magnetic field lines which come from right to left. As a result, if the charge was proton, it would move in the out-of-page direction but it is electron instead, so it will move in the onto-the-page direction. We can use the same approach for larger charged objects as well. In this case, we apply the vector equation F*⃗* = Q ∙ (v*⃗* × B*⃗*)

or its scalar equivalent F = Q ∙ v ∙ B ∙ sinθ

to calculate the magnetic force of a charged object in motion. This force is the same force we have called earlier as "the Ampere's force". This is because F = I ∙ B ∙ L ∙ sinθ = (Q/t) ∙ B ∙ (v ∙ t) ∙ sinθ = Q ∙ v ∙ B ∙ sinθ

Let's consider another example in this regard. Example: A 20 cm long current carrying wire, which is able to carry 4A of current in 10 seconds is placed between the poles of a U-shaped (horseshoe) magnet as shown in the figure. If the magnetic field produced by the magnet is 50 mT, calculate: a) Magnetic force produced on the wire b) Direction of magnetic force c) Moving velocity and direction of the wire if its weight is negligible Solution Clues: L = 20 cm = 0.2 m I = 4 A t = 10 s B = 50 mT = 0.05 T a) We can find the magnetic force using the equivalence between the Ampere's Force and the magnetic force of charges in motion. Given that the magnetic field lines lie from North to South pole of magnet (i.e. vertically down), the angle θ is 900 (that is sin θ = 1). Thus, we obtain for the magnitude of magnetic force on the wire: F = I ∙ B ∙ L ∙ sinθ = (4A) ∙ (0.05T) ∙ (0.2m) ∙ 1 = 0.04 N

b) The direction of magnetic force is found by applying the Fleming's Left Hand Rule. The palm is placed face up as the magnetic field lines lie from up to down (they must punch the palm). The four fingers are directed due left as current flows from right to left. As a result, the magnetic force (shown by the thumb) is directed outwards (out of page). The figure below gives a clearer idea on this. c) The moving direction of wire is the same as that of magnetic force as all movements are caused by forces. As for the magnitude of velocity, we use the equation F = Q ∙ v ∙ B ∙ sinθ

where F is the same magnetic force found at (a). It is equal to 0.04 N. On the other hand, the charge Q passing through the wire is Q = I ∙ t = (4A) ∙ (10s) = 40 C

Therefore, the moving velocity of wire is v = F/(Q ∙ B ∙ sinθ ) = ((0.04 N))/((40C) ∙ (0.05T) ∙ 1) = 0.02 m/s = 2 cm/s

Moving Trajectory of a Particle inside a Magnetic Field In the previous paragraph, we explained that magnetic force is always perpendicular to the moving direction of charges. (Do not confuse the moving direction of charges - which is usually determined by the shape of conducting wire - with the direction of wire's motion - which occurs in the direction of magnetic force. They are completely different things.) Given this, it is clear that the magnitude of velocity (that is the speed) of moving charges placed inside a magnetic field is constant, as in this case, the velocity has no component in the direction of magnetic force. This means the kinetic energy of charged particles inside a magnetic field is constant as well (KE = mv2/2). From the law of Work-Kinetic Energy conversion, we known that when the kinetic energy is constant, no work is done by the magnetic force on moving charges (W = ΔKE = KEf - KEi = 0). Such a situation (the displacement of charges be zero when charges are in motion) can occur only if the motion is periodic, i.e. when it repeats itself in equal time intervals. The most common periodic type of motion is the uniform circular motion. Therefore, any particle placed inside a uniform magnetic field will move in uniform circular motion, where the magnetic force will act as a centripetal force, as shown in the figure below. The radius of the curved path is obtained through the magnetic-centripetal force equivalence. Thus, we have F_mag = F_{c}p Q ∙ v ∙ B = (m ∙ v^{2})/r Q ∙ B = (m ∙ v)/r

Therefore, we obtain for the radius of circular path followed by a charged particle in a magnetic field: r = (m ∙ v)/(Q ∙ B)

where m is the mass of the charged particle. Thus, if the magnetic field is uniform (B is constant), the radius r of the particle's circular path does not change as all the other quantities in the above formula are constant as well. Given that the period of a uniform circular motion is calculated by dividing circumference by speed, we obtain for the period of rotation for a particle in circular motion inside a uniform magnetic field: T = 2πr/v = (2π ∙ (m ∙ v)/(Q ∙ B))/v = (2π ∙ m)/(Q ∙ B)

As you can see, the period of rotation is independent from the moving speed of the particle. Example: Two electric charges of equal magnitudes (4 μC each) enter inside a 40 mT uniform magnetic field as shown in the figure. Calculate: a) The type of charge Q1 and Q2 the corresponding objects contain b) The masses of the two charges c) The speeds by which charges move inside the field d) The period of rotation for each charge e) The ratio between magnetic forces F1 / F2 produced Take mproton = 1.67 × 1027 kg, melectron = 9.1 × 10-31 kg, e = 1.6 × 10-19 C. Solution a) Using the Fleming's Left Hand Rule, we find out that the charge Q1 is positive and Q2 is negative. Another way to prove this is to see the turning direction of charges. Thus, when a positive charge enters inside an onto-the-page magnetic field, the magnetic force is equal to the corresponding centripetal force, producing an anticlockwise turning effect as explained earlier, while when the charge is negative, it rotates in the opposite direction (clockwise). b) Since all values of electric charges - whether positive or negative - are multiples of elementary charge, first we find the number n of protons or electrons contained in each charge and then, we multiply them by the mass of the corresponding elementary charges e. We have Q = n ∙ e⟹n = Q/e = (4 × 10^{-6} C)/(1.6 × 10^(-19) ) = 2.5 × 10^{1}3 charges

Since Q1 = +4μC and Q2 = -4μC, we obtain for the masses m1 and m2 of the charged objects: m_{1} = n ∙ m_p = 2.5 × 10^{1}3 ∙ 1.67 × 10^(-27) kg = 4.175 × 10^{-14} kg

and m_{2} = n ∙ m_e = 2.5 × 10^{1}3 ∙ 9.1 × 10^(-31) kg = 2.275 × 10^(-17) kg

c) The speeds of each charged object are calculated using the equation r = (m ∙ v)/(Q ∙ B)

From the figure we can see that r1 = 3 cm = 3 × 10-2 m (3 units in the figure), and r2 = 1 cm = 10-2 m (1 unit in the figure). Therefore, for the positive charge Q1 we have v_{1} = (r_{1} ∙ Q_{1} ∙ B)/m_{1} = ((3 × 10^{-2} m) ∙ (4 × 10^{-6} C) ∙ (4 × 10^{-2} T))/(4.175 × 10^{-14} kg) = ├ ├ 11.497 × 10^{4} m/s┤┤ = 114 970 m/s

and for the negative charge Q2, we have v_{2} = (r_{2} ∙ Q_{2} ∙ B)/m_{2} = ((1 × 10^{-2} m) ∙ (4 × 10^{-6} C) ∙ (4 × 10^{-2} T))/(2.275 × 10^(-17) kg) = 7.033 × 10^{7} m/s = 70 330 000 m/s

d) The period of rotation for the first charge is T_{1} = (2π ∙ m_{1})/(Q_{1} ∙ B) = ((2 ∙ 3.14) ∙ (4.175 × 10^{-14} kg))/((4 × 10^{-6} C) ∙ (4 × 10^{-2} T) ) = 1.64 × 10^{-6} s

and that of the second charge is T_{2} = (2π ∙ m_{2})/(Q_{2} ∙ B) = ((2 ∙ 3.14) ∙ (2.275 × 10^(-17) kg))/((4 × 10^{-6} C) ∙ (4 × 10^{-2} T) ) = 8.93 × 10^{-10} s

e) We can calculate the magnetic forces acting on each charge in two ways: 1- Using the formula F = Q ∙ v ∙ B for each charge, or 2- Using the formula of centripetal force F = m ∙ v2 / r for each charge. Let's use the second method in this exercise. For the first charge, we have F_{1} = (m_{1} ∙ v_{1}^{2})/r_{1} = ((4.175 × 10^{-14} kg) ∙ (11.497 × 10^{4} m/s)^{2})/((3 × 10^{-2} m) ) = 1.84 × 10^{-2} N

and for the second charge, we have F_{2} = (m_{2} ∙ v_{2}^{2})/r_{2} = ((2.275 × 10^(-17) kg) ∙ (7.033 × 10^{7} m/s)^{2})/((1 × 10^{-2} m) ) = 112.5 × 10^{-1} N = 11.25 N

These results are confirmed by means of the first method. Thus, for the first magnetic force, we have F_{1} = Q_{1} ∙ v_{1} ∙ B = (4 × 10^{-6} C) ∙ (11.497 × 10^{4} m/s) ∙ (4 × 10^{-2} T) = 184 × 10^{-4} N = 1.84 × 10^{-2} N

and for the second magnetic force, we have F_{2} = Q_{2} ∙ v_{2} ∙ B = (4 × 10^{-6} C) ∙ (7.033 × 10^{7} m/s) ∙ (4 × 10^{-2} T) = 112.5 × 10^{-1} N = 11.25 N

As you see, the results are the same in both methods. Lorentz Force So far we have considered only the magnetic force caused when a charged particle enters inside a magnetic field. But in Section 14, we have seen that a charged particle produces an electric field and therefore an electric force around it. Hence, the total force produced on a charged object due to the existence of the two fields - electric and magnetic - is the sum of the corresponding forces. This overall effect is known as the Lorentz Force and its vector form is F*⃗*_Lor = F*⃗*_el+F*⃗*_mag = Q ∙ E*⃗*+Q ∙ (v*⃗* × B*⃗*)

The electric force is easier to find as it is always in the direction of electric field (when it is produced by a positive charge). As for the magnetic force, its direction is found using the Fleming's Left Hand rule as discussed earlier. Remarks! 1) In many cases (even in some textbooks), the Lorentz Force is confused with its magnetic component. People hardly realize there is also an electric component included in the Lorentz Force. This occurs because when charges are inside a strong magnetic field, the electric component of Lorentz Force is very small when compared to the corresponding magnetic component. 2) The concept of Lorentz Force is a good example why the actual chapter is called Electromagnetism. It is a demonstration of the fact that electricity and magnetism cannot exist without each other. 3) Lorentz Force exists only when charges are in motion. Stationary charges produce only an electric effect, not a magnetic one. Example: A +2mC electric charge drops at 40 cm/s between two metal plates which are 80 cm apart. The charge is light enough to fall down at terminal (uniform) velocity. The plates are connected to a 3V battery. Two bar magnets producing a magnetic field of 0.5T the opposite poles of which are facing each other close the rectangular frame as shown in the figure. a) What is the force (magnitude and direction) acting on the charge when it enters between the metal plates and magnets if the switch is OFF? b) What is the force (magnitude and direction) acting on the charge when it enters between the metal plates and magnets if the switch is ON? Solution Clues: Q = +2mC = +2 × 10-3 C d = 80 cm = 0.8 m ε = 3V B = 0.5 T v = 40 cm/s = 0.4 m/s = 4 × 10-1 m/s a) When the switch is OFF, no current flows through the circuit. As a result, the charge is under the effect of magnetic field only (we neglect the effect of gravity as the terminal velocity of the positive charge is already reached prior to entering between the poles of magnets). Since the direction of velocity is perpendicular to magnetic field lines that extend from N to S pole (sin 900 = 1), we obtain for the magnitude of magnetic force F_m = Q ∙ v ∙ B = (2 × 10^{-3} C) ∙ (4 × 10^{-1} m/s) ∙ (0.5T) = 4 × 10^{-4} N

The direction of this magnetic force is due left. This is found by applying the Fleming's Left Hand Rule (the palm is punched by magnetic field lines, the four fingers show the direction of positive charge, i.e. downwards and thumb shows the direction of force). b) When the switch turns ON, an electric field is produced besides the magnetic field. The direction of this electric field will be from left to right. This is because the left plate is charged positively as it is connected to the positive terminal of battery. (We know that the direction of electric field is from positive to negative charges). Therefore, the corresponding electric force is from left to right as well. It is in opposite direction to the magnetic field found earlier. The magnitude of this electric force is F_e = Q ∙ E = (Q ∙ ε)/d = ((2 × 10^{-3} C) ∙ (3V))/((0.8m) ) = 7.5 × 10^{-3} N

Therefore, the total (net) force - which here represents the Lorentz Force - is F_Lor = F_e-F_m = 7.5 × 10^{-3} N-4 × 10^{-4} N = 7.5 × 10^{-3} N-0.4 × 10^{-3} N = 7.5 × 10^{-3} N

Therefore, the Lorentz Force is 7.5mN in the direction of electric force (from left to right). Summary The magnetic force F1 acting on each charge due to their directed motion in the conducting wire is calculated by dividing the total magnetic force Ftot by the number of charges n flowing in the entire length L of the wire. Mathematically, we have: F_{1} = F_tot/n

Giving that the current I is I = ∆Q/∆t

where ΔQ is the charge flowing through the wire in the time interval Δt and L = v ∙ ∆t

where v is the velocity of moving charges throughout the wire during this interval, we obtain for the magnetic force acting on an elementary charge in motion F_{1} = = e ∙ B ∙ v

where e = ∆Q/n

The vector form of the above equation is F*⃗*_{1} = e ∙ (v*⃗* × B*⃗*)

and the direction of magnetic force is found by using the Fleming's Left Hand Rule. Not always the direction of particles motion is perpendicular to the magnetic field lines. When these two vectors form another angle θ to each other, we have to consider this angle as well. The formula of magnetic force for an elementary electric charge in such conditions therefore becomes F_{1} = e ∙ v ∙ B ∙ sinθ

However, the force vector will still be perpendicular to the plane of the other two vectors (v and B). We can use the same approach for larger charged objects as well. In this case, we apply the vector equation F*⃗* = Q ∙ (v*⃗* × B*⃗*)

or its scalar equivalent F = Q ∙ v ∙ B ∙ sinθ

to calculate the magnetic force of a charged object in motion. This force is the same force we have called earlier as "the Ampere's force". Any particle placed inside a uniform magnetic field will move in uniform circular motion, where the magnetic force will act as a centripetal force. The radius of the resulting curved path is obtained by analyzing the magnetic-centripetal force equivalence, i.e. F_mag = F_{c}p Q ∙ v ∙ B = (m ∙ v^{2})/r r = (m ∙ v)/(Q ∙ B)

The period of rotation for a particle in circular motion inside a uniform magnetic field is: T = 2πr/v

or T = (2π ∙ m)/(Q ∙ B)

The total force produced on a charged object due to the existence of the two fields - electric and magnetic - is the sum of the corresponding forces. This overall effect is known as the Lorentz Force and its vector form is F*⃗*_Lor = F*⃗*_el+F*⃗*_mag

or F*⃗*_Lor = Q ∙ E*⃗*+Q ∙ (v*⃗* × B*⃗*)

The electric force is easier to find as it is always in the direction of electric field (when it is produced by a positive charge). As for the magnetic force, its direction is found using the Fleming's Left Hand rule. Lorentz Force is a demonstration of the fact that electricity and magnetism cannot exist without each other. Lorentz Force exists only when charges are in motion. Stationary charges produce only an electric effect, not a magnetic one. **1)** A single proton having a velocity of 2 × 105 m/s enters a 200mT uniform magnetic field at 300 to the field lines as shown in the figure.

What is the magnitude and direction of magnetic force produced on the proton? Take cos 300 = 0.866 and sin 300 = 0.500 if necessary. The magnitude of an elementary charge is 1.6 × 10-19 C.

- 3.2 × 10-15 N out of page
- 5.54 × 10-15 N out of page
- 3.2 × 10-15 N onto the page
- 5.54 × 10-15 N onto the page

**Correct Answer: C**

**2)** An object having a charge of +6.4 × 10-17 C enters a 2μT magnetic field at 105 m/s as shown in the figure.

What is the radius of curvature of the charge's trajectory inside the magnetic field?

(Take mp = 1.67 × 10-27 kg, me = 9.1 × 10-31 kg and e = 1.6 × 10-19 C if necessary).

- 0.29 m
- 400 m
- 1.39 m
- 557 m

**Correct Answer: D**

**3)** What is the magnitude of the net force acting on a +3μC charge when it enters at 2000 km/s inside normal to the frame obtained by a parallel plate capacitor and two bar magnets that produce a 5 mT magnetic field as shown in the figure? The capacitor plates are 2 cm apart and there is a potential difference of 100 V between them.

- 0.015 N
- 0.030 N
- 0.045 N
- 0.060 N

**Correct Answer: C**

We hope you found this Physics tutorial "Magnetic Force on a Wire Moving Inside a Magnetic Field. Lorentz Force" useful. If you thought the guide useful, it would be great if you could spare the time to rate this tutorial and/or share on social media, this helps us identify popular tutorials and calculators and expand our free learning resources to support our users around the world have free access to expand their knowledge of physics and other disciplines. In our next tutorial, we expand your insight and knowledge of Magnetism with our Physics tutorial on Magnetic Moment .

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