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Unit 16 Magnetism 16.1 Permanent magnets 16.2 The magnetic force on moving charge 16.3 The motion of charged particles in a magnetic field 16.4 The magnetic force exerted on a current-carrying wire 16.5 Current loops and magnetic torque 16.6 Biot and Savart’s law 16.1 Permanent magnets A bar magnet can attract another magnet or repel it, depending on which ends of the magnets are brought together. One end of a magnet is referred to as its north pole; the other end is its south pole. The rule for whether two magnets attract or repel each other: opposites attract; likes repel. Breaking a magnet in half results in the appearance of two new poles on either side of the break. This behavior is fundamentally different from that in electricity, where the two types of charge can exist separately. We saw a visual indication of the electric field E of a point charge using grass seed suspended in oil. Similarly, the magnetic field B can be visualized using small iron filings sprinkled onto a smooth surface. The filings are bunched together near the poles of the magnets. This is where the magnetic field is most intense. The direction of the magnetic field, B, at a given location is the direction in which the north pole of a compass points when placed at that location. In general, magnetic field lines exit from the north pole of a magnet and enter at the south pole. 1 16.2 The magnetic force on moving charge The magnetic force depends on several factors: • The charge of the particle, q; • The speed of the particle, v; • The magnitude of the magnetic field, B; • The angle between the velocity vector and the magnetic field vector, θ. The mathematical relation of them in vector form is F = q (v × B ) . One can rewrite it as a scalar expression, e.g. F = qvB sin θ . The maximum force is obtained when θ = 90o. The force vanishes when θ = 0o. Now we define the magnetic field B as B = F . The SI unit is 1 tesla = 1 T = 1 N/(A⋅m). The tesla is a fairly qv sin θ large unit of magnetic strength, especially when compared with the magnetic field at the surface of the Earth, which is roughly 5.0 × 10 −5 T . Thus, another commonly used unit of magnetism is the gauss (G), defined as follows: 1G = 10 −4 T . In terms of the 2 gauss, the Earth’s magnetic field on the surface of the Earth is approximately 0.5 G. A bar magnet has a magnetic field of roughly 100 G. Remark: Magnetic field lines never cross one another. As the direction in which a compass points at any given location is the direction of the magnetic field at that point. Since a compass can point in one direction, there must be only one direction for the field B. If field lines were to cross, however, there would be two directions for B at the crossing point, and this is not allowed. Example Particle 1, with a charge q1 = 3.60 µC and a speed v1 = 862 m/s travels at right angles to a uniform magnetic field. The magnetic force it experiences is 4.25 × 10−3 N . Particle 2, with a charge q2 = 53.0 µC and a speed = v2 1.30 × 103 m / s moves at an angle of 55.0o relative to the same magnetic field. Find (a) the strength of the magnetic field and (b) the magnitude of the magnetic force exerted on particle 2. Answer Apply the formula: F = qvB sin θ with θ = 90o. One obtains B = 4.25 × 10−3 N F = = 1.37 T . qv sin θ (3.60 × 10−6 C )(862 m / s ) sin 90o Apply the formula: F = qvB sin θ with θ = 55.0o. One obtains F= (53.0 × 10−6 )(1.30 × 103 m / s ) (1.37 T ) sin 55.0o = 0.0773 N . The direction of the magnetic force is given by the magnetic force right-hand rule (RHR), which states as follows. To find the direction of magnetic force on a positive charge, start by pointing the fingers of your right hand in the direction of the velocity, v. Now curl your fingers forward the direction of B. Your thumb points in the direction of F. 3 If the charge is negative, the force points opposite to the direction of your thumb. Note that the magnetic force F points in a direction that is perpendicular to both B and the charge velocity v. As an example, the direction of a magnetic force F can be indicated by the magnetic force RHR – extending our fingers to the right (v) and then curling them into the page (B) – we see that the magnetic force exerted on this particle is upward, as indicated. If the charge is negative, the direction of F is reversed. Example Three particles travel through a region of space where the magnetic field is out of the page as shown in figure. For each of the three particles, state whether the particle’s charge is positive, negative, or zero. Answer Use the RHR to obtain the following. Particle 1: negative; particle 2: zero and particle 3: positive. 4 16.3 The motion of charged particles in a magnetic field Figures (a) and (b) shows the motion of a positive charged particle moving under electric field and magnetic field respectively. Remarks: 1. As the direction of velocity and the direction of magnetic force are perpendicular to each other, the work done by the magnetic force is always zero. 2. A charged particle with a velocity v that is perpendicular to the magnetic field moves in a circular path. The magnetic force acts as the centripetal force, e.g. Fcp = Fcp = qvB . Hence path, r = mv 2 and r mv 2 = qvB gives the radius of circular r mv 2 mv . On the other hand, Fcp = = mω 2 r = qvB , qB r we can write mω = qB , as v = rω . Plugging in the relation 3. ω= 2π 2π , we have m = qB , and the period T is given by T T T= 2πm . qB Helical motion is a combination of linear motion and circular motion. 5 Example In a device called a velocity selector, charged particles move through a region of space with both an electric and a magnetic field. If the speed of the particle has a particular value, the net force acting on it is zero. Assume that a positively charged particle moves in the positive x direction, and the electric field is in the positive y direction. Should the magnetic field be in (a) the positive z direction, (b) the negative y direction, or (c) the negative z direction in order to give zero net force? Answer The force exerted by the electric field is in the positive y direction; hence, the magnetic force must be in the negative y direction if it is to cancel the electric force. If we simply try the three possible directions for B one at a time, applying the magnetic force RHR in each case, we find that only a magnetic field along the positive z axis gives rise to a force in the negative y direction, as desired. The answer is (a). Example Two isotopes of uranium, 235 U ( 3.90 × 10−25 kg ) and 238 U ( 3.95 × 10−25 kg ), are sent into a mass spectrometer with a speed of 1.05 × 105 m / s . Given that each isotope is singly ionized, and that the strength of the magnetic field is 0.750 T, what is the distance d between the two isotopes after they complete half a circular orbit? Answer The isotopes are singly ionized, which means that a single electron has been removed from each atom. And the isotopes are now having charge of = e 1.60 × 10−19 C . 6 From the relation mv 2 rqB mv , and thus r = . = qvB , we obtain v = m r qB r= 235 mv (3.90 × 10−25 kg )(1.05 × 105 m / s ) = = 34.1 cm . qB (1.60 × 10−19 C )(0.750 T ) r= 238 mv (3.95 × 10−25 kg )(1.05 × 105 m / s ) = = 34.6 cm . qB (1.60 × 10−19 C )(0.750 T ) The separation between the isotopes: d = 2r238 − 2r235 = 2(34.6 cm − 34.1 cm) = 1cm . 16.4 The magnetic force exerted on a current-carrying wire As a charged particle experiences a force when it moves across magnetic field lines. The same thing happens when a wire carries current on it. Consider a straight segment of length L of a wire with a current I flowing from left to right, presents in a magnetic field B, as shown in figure. If the conducting charges move through the wire with an average speed v, the time required for them to move from one end of the wire segment to the other is ∆t = L / v . The amount of charge that flows through the wire in this time is q = I ∆t = IL / v . Therefore, the force exerted on the wire is I L F = qvB sin θ = vB sin θ . v Hence, we have F = ILB sin θ . Maximum force occurs when the current is perpendicular to the magnetic field (θ = 90o) and is zero if the current is in the same direction as B (θ = 0o). The direction of the magnetic force is given by the RHR, where the direction of charge velocity v is now the direction of current I. Example When the switch is closed in the circuit, the wire between the poles of the horseshoe magnet deflects downward. Is the left end of the magnet (a) a north magnet pole of (b) a south magnetic pole? 7 Answer Once the switch is closed, the current in the wire is into the page, as shown in the right figure. Applying the magnetic force RHR, we see that the magnetic field must point from left to right in order for the force to be downward. Since magnetic field lines leave from north poles and enter at south poles, it follows that the left end of the magnet must be a north magnetic pole. The answer is (a). 16.5 Current loops and magnetic torque A torque is experienced by a current loop as shown in figure. If we imagine an axis of rotation through the center of the loop, at the point O, it is clear that the forces exert a torque that tends to rotate the loop clockwise. w 2 w 2 τ = ( IhB) + ( IhB) = IB(hw) = IAB , Plug in the area of the rectangular loop, A = hw. We have τ = IAB . If the plane of loop makes an angle with the magnetic field, we have τ = IAB sin θ . For the case that has n turns in a general loops, we have τ = nIAB sin θ . Applications of the magnetic torque are electric motors and galvanometer. 8 16.6 Biot and Savart’s law B −field E −Electric field Charge Current Charge Current Recall that in the Coulomb’s law, we have, say, a point charge q gives an electric field at P and E = 1 q rˆ . For a charge 4πε 0 r 2 distribution, the electric field is at P due to dq is dE = dq dE r 1 dq rˆ . 4πε 0 r 2 Now, for a magnetic field, we have the law of Biot and Savart dB = µ 0 I ds sin θ , 4π r2 where µ0 = 4π × 10−7 T ⋅ m / A = 4π × 10−7 H / m I ( 1 Henry = 1H = 1T ⋅ m / A ) and µ 0 is called the 2 ds permeability constant. In vector form, we µ 0 I ds × rˆ have dB = . Comparing with the Coulomb’s 4π r 2 Magnetic field θ r r̂ I 1 law, we realize that both are 2 laws. r Example Find the magnetic field at the center of a coil which carries a steady current I. Answer The magnetic field through the center of coil is the superposition of the magnetic field due to current segment. Hence, B = ∑ As = B r is a µ0 I ds sin θ . 4π r2 constant and θ = 90o, we can write µ0 I µ0 I µ0 I . = ds = (2 r ) π ∑ 2r 4π r 2 4π r 2 If the coil has N turns, the magnetic field is given by B = µ0 NI 2r . 9 .× P Remark: If the solenoid has N turns, length l and carries a current I, the magnetic field B at a point O on the axis near the center of the solenoid is found to be = B µ0 NI = µ0 nI , l where n is the number of turns per unit length of coil. If the coil is infinite long or long enough that its length is ten times the diameter, the magnetic field at the end of coil is half that at the center of coil, e.g. B = µ0 nI 2 . Example (Challenging) Find the magnetic field B at a point P, a distance z from a long current wire, as shown in the figure. Answer From Biot-Savart’s law, we have µ0 I dl × rˆ , where dl = B= φ dl cos θ . × rˆ dl sin = 2 ∫ 4π r As l = z tan θ , we have dl = ( z / cos 2 θ ) dθ . µI We can write B = 0 4π ∫ π 2 − π 2 dl cos θ . r2 P z 1 cos θ Now, z / r = cos θ ⇒ = gives r z µI B= 0 4π = π z cos 2 θ 2 ∫−π2 cos2 θ z 2 cos θ dθ I r θ O φ l dl π µ0 I π2 µ0 I µ0 I 2 d cos sin θ θ θ = = π π ∫ − 4π z − 2 4π z 2π z 2 Hence, the magnetic field at P due to a long wire of current is given by B = µ0 I . 2π z 10 Remark 1: The magnitude of the magnetic field 1 m from a long, straight wire carrying a current of 1 A is B= µ0 I (4π × 10−7 T ⋅ m / A)(1 A) = = 2 × 10−7 T . 2π z 2π (1 m) This is a weak field, less than one hundredth the strength of the Earth’s magnetic field. Remark 2 The magnetic field shown in the figure is due to the horizontal, current-carrying wire. The current in the wire flows to the left. The following states the use of magnetic field right-hand rule which points the direction of current: If you point the thumb of your right hand along the wire to the left your fingers curl into the page above the wire and out of the page below the wire. Thus, the current flows to the left. 11