Magnetic Force - Force on a Current Carrying Conductor

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(Felectric+Fmagnetic ). Mathematically,

\(F=q[E(r)+v\times B(r)]\)

This force is termed as the Lorentz force. The interaction between the electric field and the magnetic field has the following features:

The magnetic field depends upon the charge of the particle, the velocity of the particle and the magnetic field in which it is placed. The direction of the magnetic force is opposite to that of a positive charge.

The magnitude of the force is calculated by including the cross product of velocity and the magnetic field, given by q [ v × B ]. The resultant force is thus perpendicular to the direction of the velocity and the magnetic field, the direction of the magnetic field is predicted by the right hand thumb rule.

In the case of static charges, the total magnetic force is zero.

Magnetic force on a current carrying conductor

In this section, we will learn about the force due to magnetic field in a straight current carrying rod. We consider a rod of uniform length l and cross sectional area A. In the conducting rod, let the number density of mobile electrons be given by n. Then the total number of charge carriers can be given by nAI, where I is the steady current in the rod. The drift velocity of each mobile carrier is assumed to be given as vd. When the conducting rod is placed in an external magnetic field of magnitude B, the force applied on the mobile charges or the electrons can be given as

\(F=(nAI)qvd\times B\)

Where q is the value of charge on the mobile carrier. As nqvd is also the current density j and A×|nqvd| is the current I through the conductor, then we can write,

\(F=[(nqevd)AI]\times B=[jAI]\times B=Il\times B\)

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

Stay tuned with Byju’s to learn more about magnetic force, drift velocity and much more.

Practise This Question

A particle of charge +q and mass m moving under the influence of a uniform electric field E^i and a uniform magnetic field B^k follows trajectory from P to Q as shown in figure. The velocities at P and Q are v^i and 2v^j respectively. Which of the following statement(s) is/are correct