Magnetic force is the consequence of electromagnetic force. It is caused due to the motion of charges. A student has learned that moving charges surround themselves with a magnetic field. With this concept, the magnetic force will be described as a force that arises due to interacting magnetic fields. In this topic, we will discuss the Magnetic Force Formula with examples. Let us learn the concept!

**What is a Magnetic Force?**

If we place a point charge q in the presence of both a magnitude field given by magnitude B(r) and an electric field given by a magnitude E(r). Then the total force on the electric charge q can be written as the sum of the electric force and the magnetic force acting on the object as: \((F_{Electric} + F_{Magnetic} )\)

The magnetic force between two moving charges is defined as the effect exerted upon by a magnetic field created by the other. The value of the magnetic force depends on how much charge is in the motion.

Mathematically, we can write magnetic force as: \(F = q[E(r) + v \times B(r)\)

This above force is termed as the Lorentz Force. It is the combined electric and magnetic force on a point charge due to the electromagnetic fields. The interaction between these two fields have the following features:

- The magnetic force depends upon the charge of the particle and the velocity of the particle as well as on its magnetic field. The direction of the magnetic force is the opposite of that of the positive charge.
- The magnitude of the force can be calculated by the cross product of velocity and the magnetic field. This is given by \(q [ v \times B ]\)
- The resultant force is always perpendicular to the direction of the velocity and the magnetic field.
- For the static charges, the total magnetic force is zero.

**Magnetic Force on a Conductor Carrying Current**

Now, we will discuss the force due to the magnetic field in a rod that is straight and carrying current. We can consider a rod of uniform length l and cross-sectional area A. In this rod, let the number density of mobile electrons is n.

Then the total number of charge carriers is as: q = nAI, where I is the steady current in the rod.

The drift velocity is v_d. If the rod is placed in an external magnetic field B, then the force applied on the mobile charges or the electrons will be as: \(F = nAI \times qv_d \times B\). Where q is the value of charge on the mobile carrier.

As \(nqv_d\) is also the current density \(j and A \times (nqv_d\)) is the current I through the conductor, then we can write:

\(F = (nqev_d)AI \times B\)

\(F = (jAI) \times B\)

\(F = I\times l \times B\)

Where I is the vector of magnitude equal to the length of the conducting rod.

If \(\theta\) is the angle between length and magnetic field vectors in radians, then \(F = I l B Sin\theta \hat{n}\). Where \(\hat{n}\) is the cross product direction vector, which is a unitless quantity.

## Solved Examples for Magnetic Force Formula

Q.1: The direction of the current in a conducting wire carrying a current of 6.00 A through a uniform magnetic field 2.20T is from the left to right of the screen. The direction of the magnetic field is upward-left, with an angle of\( θ = \frac{3 \pi }{4}\) radians from the current direction.

Find out the magnitude and direction for the magnetic force acting on the section of wire of length 0.100 m using Magnetic Force Formula

Solution: Given,

I= 6.0 A

L= 0.1 m

B= 2.2 T

\(\theta = \frac{3 \pi }{4} radians\)

n is the cross product direction vector (unitless)

Substituting the values in formula,

\(F = I l B Sin\theta\)

we get, \(F = (6.00) \times (0.100) \times (2.20) \times sin(\frac{3 \pi }{4}) = 0.933 \;kgms^{-2}\\ \)

Therefore, the magnitude of the force on the 0.100 m section of wire has a magnitude of 0.933 N. We will use “right-hand rule” to find the direction of the force vector. Therefore, the direction of the force vector is out of the page.

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