PHYSICS

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of electric charge; likewise, a wire with current passing through it creates a magnetic field due to the charge moving in the wire. Bar magnets also produce a magnetic field. Let us define a magnetic field vector lJ (sometimes called the magnetic induction) at some point in space in terms of a magnetic force would be exerted on an appropriate object - a charged particle moving with a velocity V. Magnetic force (Lorenz force) acting on a particle with charge q depends on the velocity V of the charge and is equal to PM = q [V x B] (18.1) where [V x Bj is the cross product of vectors V and lJ (fig. 18.1). This force is always perpendicular to both vectors V and s. Fig. 18.1. The Lorenz force PM acting on a charged particle with charge q moving with a velocity V in the presence of a magnetic field lJ _ F M F M i3 _v Fig. 18.2. The Lorenz force: a the directions of the magnetic force PM' velocity V of charged particle, and magnetic field s, b - the right-hand rule for the determining the direction of the The right-hand rule for the determination the direction of the Lorenz force (fig. 18.2): you point the four fingers of your right hand along the direction of V, and then turn them until they point along the direction of lJ,. the thumb then points in the direction of PM' 132 a b ",It ;~~; (~ ~ :J~ 'I, The unit of the magnetic field is the tesla (1 T = 1 N/Am). For a large magnetic field, the commonly used subunit is the gauss (I T = 10 4 G), e.g., the magnetic field of the Earth is about 0.5 G, while the magnetic field of a large electromagnet is about 1 T. Example. A proton moves with a speed of 8.10 6 rn/s along the x axis. It enters a region where there is a field of magnitude 2.5 T, directed at an angle of 60° to the x axis. Calculate the initial magnetic force and acceleration of the proton. Solution. From Equation (18.1) , we get: F = q Vbsine = (1.6.10- 19 C)(8·1 0 6 m/s)(2,5 T)(sin60o) = 2.77-10- 12 N. 18.2. THE MASS SPECTROMETRY Consider the case of positively charged particle moving in a uniform external magnetic field with its initial velocity vector perpendicular to the field. Let us assume that the magnetic field is into the page. Charged particle moves in a circle whose plane is perpendicular to the magnetic field. As the force PM deflects the particle, the directions of V and PM change continuously. Therefore the force PM is a centripetal force. From Newton's second law, we find that or F = qVB = mV 2 r = mV qB r (18.2) (18.3) The mass spectrometer is an instrument that separates atomic and molecular ions according to their mass-to-charge ratio. A beam of ions first passes through a velocity selector and then enters a uniform magnetic field (f3) directed into the paper (fig. 18.3). Upon entering the magnetic field, the ions move in a semicircle of radius r before striking a photograph plate at P. From Equation (18.3) we can express the ratio m/o as m ·I q rB V (18.4) 133
x x x x x p -, x x x X B x\ \ x x x x x Fig. 18.3. A mass spectro X '~ X X X X x: , x x x meter which consists of llHHI x x x x x x,/ x x x E-r - --- --- _.... a velocity selector and xV x x x x x x a source of uniform XXXXXXXlxxxxx magnetic field Velosity selector x x x x x Bo, in Therefore, one can determine m/q by measuring the radius of curvature and knowing the field B. Example. A proton is moving in a circular orbit of radius 14 em in a uniform magnetic field of magnitude 0.35 T directed perpendicular to the velocity of the proton. Find the orbital speed of the proton. Solution. From Equation (18.3), we get V = qBr = (1.6 ·lO'19)(0.35T)(l4 ·10'2 m) = 4.69 . 10 6 m / s m 1.67 . 10'27 kg 18.3. AMPERE'S LAW Let us consider a straight segment of wire of length I, carrying a current I in an uniform external magnetic field (ii) (fig. 18.4). The total magnetic force on the wire is dF = I [dT x B] (18.5) where dT is a vector in the direction of the current I. Thus the force on a differential length dl of current carrying wire is the vector cross product of dT and the magnetic field fJ at dT. 18.4. MAGNETIC FIELD OF A THIN STRAIGHT CONDUCTOR Consider the case of an infinitely long straight wire. The total magnetic field at the point located at a distance a from the wire is given as: B = flof (18.6) 2na Example. Calculate the magnetic field of a long, straight wire carrying a current of 5 A, at a distance of 4 cm from the wire. Solution. From equation (18.6), we get: B = Jiol = 4n ·10'7 N / ~ 2 ·5A = 2.5' 1O-5T. Zna 2n . 4 . 10" m 18.5. FARADAY'S LAW Let us consider a loop of wire connected to a galvanometer; if a magnet is moved toward or away from the loop, galvanometer deflects. It means that a current is set in the circuit as long as there is relative motion between the magnet and the loop. Such a current is produced by an induced electromotive force (E.M.F.) in a loop. The E.M.F. induced in a circuit is directly proportional to the time rate of change of magnetic flux (if]M) through the circuit. This statement is known as Faraday's Law of Induction: E = _ d
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