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Transcript
Chapter IV Magnetic Force Magnetic Field: A magnetic field is generated when electric charge carriers such as electrons move through space or within an electrical conductor. The symbol H refers to the strength of magnetic field. Figure (4.1): Magnetic field of an ideal cylindrical magnet with its axis of symmetry inside the image plane. The magnetic field is represented by magnetic field lines, which show the direction of the field at different points. Magnetic Induction: The Magnetic induction (B) is defined as follows: B 0 H Where: 0 is the permittivity and H is the magnetic field strength. The magnetic induction (B) is measured in Tesla (T) or in Newton/ Ampere. Meter (N/A. m). Magnetic Flux: The Magnetic flux (φ) is defined as follows: B A Where: B is the magnetic induction and A is the area The Magnetic Field Created by a Long Current-Carrying Wire ---- The Magnetic Force Between Two Parallel Conductors Consider two long, straight, parallel wires separated by a distance a and carrying currents I 1 and I 2 in the same direction as in the Figure 4.3. Let’s determine the force exerted on one wire due to the magnetic field set up by the other wire. Wire 2, which carries a current I2 and is identified arbitrarily as the source wire, creates a magnetic field B 2 at the location of wire 1, the test wire. The magnitude of this magnetic field is the same at all points on wire 1. The direction of B 2 is perpendicular to 1 ℓ wire 1 as shown in the Figure. I1 According to the following equation, B2 FB I LxB 2 F1 a the magnetic force on a length ℓ of wire 1 is F 1 = I1 x B 2 . Figure 4.2 Because ℓ is perpendicular to B 2 , in this situation, the magnitude of F 1 is F 1 = I1 B2 . Because the magnitude of B 2 is given by B2 0 I 2 a I2 Therefore, I F1 I1 B2 I1 0 2 2a I I F1 0 1 2 2a Example (4.1): Two parallel conducting plates, a current of 10 A passes through one of them and a current of 20 A passes through in the other one, calculate the following: The force per unit length (F/ℓ). Define the point at which B1= B2 , if the currents (I1 and I2) in the same directions. Define the point point at which B1= B2 , if the currents (I1 and I2) in opposite directions. I2=20 A I1=10 A Solution: The force per unit length (F/ℓ) F 0 I 1 I 2 2 a F 4 x x107 x10 x 20 = 1.33x10-4 N/m 2 2 x x 30 x10 30 cm The point at which B1= B2 , if the currents (I1 and I2) in the same directions I1 I2 x 0.3 x 10 20 x 0.3 x 20x 10 x 0.3 x 20x 10x 3 30x 3 x 3 0.1 m 30 The point at which B1= B2 , if the currents (I1 and I2) in opposite directions I1 I2 x 0.3 x 20x 10 x 0.3 x 10 20 x 0.3 x 20x 10x 3 10x 3 x 3 0.3 m 10 Magnetic Force: The existence of a magnetic field B at some point in space can be determined by measuring the magnetic force FB exerted on an appropriate test particle placed at that point. This process is the same in defining the electric field. If we perform such an experiment by placing a particle with charge q in the magnetic field, it is found the following results that are similar to those for experiments on electric forces: The magnetic force is proportional to the charge q of the particle. The magnetic force on a negative charge is directed opposite to the force on a positive charge moving in the same direction. The magnetic force is proportional to the magnitude of the magnetic field vector B . Also, the following results, which are totally different from those for experiments: on electric forces should taken in our consideration: The magnetic force is proportional to the speed v of the particle. If the velocity vector makes an angle θ with the magnetic field, the magnitude of the magnetic force is proportional to sin θ. When a charged particle moves parallel to the magnetic field vector, the magnetic force on the charge is zero. When a charged particle moves in a direction not parallel to the magnetic field vector, the magnetic force acts in a direction perpendicular to both v and B that is, the magnetic force is perpendicular to the plane formed by v and B . F This relation can be written as follows: F q v x B B The magnitude of the magnetic force on a charged particle is F qvB sin θ v So: F FB is zero when v is parallel or antiparallel to B (θ = 00 and 1800) and maximum when v is perpendicular to B (θ = 900). Let’s compare the important differences between the electric and magnetic versions of the particle in a field model: The electric force vector is along the direction of the electric field, whereas the magnetic force vector is perpendicular to the magnetic field. The electric force acts on a charged particle regardless of whether the particle is moving, whereas the magnetic force acts on a charged particle only when the particle is in motion. The electric force does work in displacing a charged particle, whereas the magnetic force associated with a steady magnetic field does no work when a particle is displaced because the force is perpendicular to the displacement of its point of application. The SI unit of magnetic field is the newton per coulomb-meter per second, which is called the Tesla (T): 1T 1 N C .m sec Example (4.2): An electron in an old-style television picture tube moves toward the front of the tube with a speed of 8x106 m/s along the or 1T 1 N A. m x axis. Surrounding the neck of the tube z are coils of wire that create a magnetic field of magnitude 0.025 T, directed at an -e angle of 600 to the x axis and lying in the y v xy plane. Calculate the magnetic force on the electron. x F 600 B Solution: FB = |q|vB sin θ = (1.6x1019 C) (8x106 m/s) (0.025 T)(sin 600) = 2.8x 10-14 N Motion of a Charged Particle in a Uniform Magnetic Field The particle moves in a circle because the magnetic force F B is perpendicular to v and B and has a constant magnitude qvB. r q FB As Figure 4.2 illustrates, the rotation is v + counterclockwise for a positive charge in + FB a magnetic field directed into the page. If q FB v q were negative, the rotation would be + clockwise. We use the particle under a net v q force model to write Newton’s second law for the particle: F F B ma Because the particle moves in a circle, we also model it as a particle in uniform circular motion and we replace the acceleration with centripetal acceleration: FB qvB mv 2 r This expression leads to the following equation for the radius of the circular path: r mv qB That is, the radius of the path is proportional to the linear momentum mv of the particle and inversely proportional to the magnitude of the charge on the particle and to the magnitude of the magnetic field. The angular speed of the particle is expressed as follow: v qB r m The period of the motion (the time interval the particle requires to complete one revolution) is equal to the circumference of the circle divided by the speed of the particle: T 2 2r 2m v qB These results show that the angular speed of the particle and the period of the circular motion do not depend on the speed of the particle or on the radius of the orbit. The angular speed v is often referred to as the cyclotron frequency because charged particles circulate at this angular frequency in the type of accelerator called a cyclotron. Example (4.3): A proton is moving in a circular orbit of radius 14 cm in a uniform 0.35-T magnetic field perpendicular to the velocity of the proton. Find the speed of the proton. Solution: v Substitute numerical values: qBr mP v q B r 1.6 x10 19 x 0.35 x 0.14 mP 1.67 x10 27 1.67 3 10227 kg V= 4.7x106 m/s Example (4.4): An electron moves in circular path with radius of 2x10-5 cm under the effect of magnetic field with induction of 1.5x10-3 Wb/m2. Calculate the following: Qe= 1.6x10-19 C and me= 9.11x10-31 kg Solution: The magnetic force (F) F Q v B 1.6 x1019 x 52.68 x1.5x103 1.26 x1020 N The electron velocity (v) eBr 1.6 x1019 x1.5 x103 x 2 x105 x102 m 9.11x10 31 v 52.68 m / sec v The angular velocity of electron (ω) eB 1.6 x1019 x1.5 x103 m 9.11x10 31 263x109 rad / sec .