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www.GOALias.blogspot.comfield vanish (become zero) if velocity and magnetic field are parallelor anti-parallel. The force acts in a (sideways) direction perpendicularto both the velocity and the magnetic field.Its direction is given by the screw rule orright hand rule for vector (or cross) productas illustrated in Fig. 4.2.(iii) The magnetic force is zero if charge is notmoving (as then |v|= 0). Only a movingcharge feels the magnetic force.The expression for the magnetic force helpsus to define the unit of the magnetic field, ifone takes q, F and v, all to be unity in the forceequation F = q [ v × B] =q v B sin θ ˆn , where θis the angle between v and B [see Fig. 4.2 (a)].The magnitude of magnetic field B is 1 SI unit,when the force acting on a unit charge (1 C),moving perpendicular to B with a speed 1m/s,is one newton.Dimensionally, we have [B] = [F/qv] and the unitof B are Newton second / (coulomb metre). Thisunit is called tesla (T ) named after Nikola Tesla(1856 – 1943). Tesla is a rather large unit. A smaller unit (non-SI) calledgauss (=10 –4 tesla) is also often used. The earth’s magnetic field is about3.6 × 10 –5 T. Table 4.1 lists magnetic fields over a wide range in theuniverse.Moving Charges andMagnetismFIGURE 4.2 The direction of the magneticforce acting on a charged particle. (a) Theforce on a positively charged particle withvelocity v and making an angle θ with themagnetic field B is given by the right-handrule. (b) A moving charged particle q isdeflected in an opposite sense to –q in thepresence of magnetic field.TABLE 4.1 ORDER OF MAGNITUDES OF MAGNETIC FIELDS IN A VARIETY OF PHYSICAL SITUATIONSPhysical situationMagnitude of B (in tesla)Surface of a neutron star 10 8Typical large field in a laboratory 1Near a small bar magnet 10 –2On the earth’s surface 10 –5Human nerve fibre 10 –10Interstellar space 10 –124.2.3 Magnetic force on a current-carrying conductorWe can extend the analysis for force due to magnetic field on a singlemoving charge to a straight rod carrying current. Consider a rod of auniform cross-sectional area A and length l. We shall assume one kindof mobile carriers as in a conductor (here electrons). Let the numberdensity of these mobile charge carriers in it be n. Then the total numberof mobile charge carriers in it is nAl. For a steady current I in thisconducting rod, we may assume that each mobile carrier has an average135