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Lecture 16 Magnetism (3)
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History 1819 Hans Christian Oersted discovered that a compass needle was deflected by a current carrying wire Then in 1920s Jean-Baptiste Biot and Felix Savart performed experiments to determine the force exerted on a compass by a current carrying wire
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Magnetic Field of a long straight wire I=0 I
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Biot & Savart’s Results dB the magnetic field produced by a small section of wire ds a vector the length of the small section of wire in the direction of the current r the positional vector from the section of wire to where the magnetic field is measured I the current in the wire angle between ds & r r ds dB
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Biot & Savart’s Results dB perpendicular to ds |dB| inversely proportional to |r| 2 |dB| proportional to current I |dB| proportional to |ds| |dB| proportional to sin
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Biot–Savart Law All these results could be summarised by one “Law” Putting in the constant Where 0 is the permeablity of free space
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Magnetic Field from Biot-Savart Law B = dB 1 +dB 2 +…+dB i I.e. B = dB r1r1 ds 1 dB 1 r2r2 ds 2 ds i dB i riri dB 2 We can use the Biot-Savart law to calculate the magnetic field due to any current carrying wire
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Magnetic Field due to Currents The passage of a steady current in a wire produces a magnetic field around the wire. Field form concentric lines around the wire Direction of the field given by the right hand rule. If the wire is grasped in the right hand with the thumb in the direction of the current, the fingers will curl in the direction of the field. Magnitude of the field I
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Magnitude of the field I r B o called the permeability of free space
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[Q]: The two wires in the figure below carry currents of 3.00A and 5.00A in the direction indicated. Find the direction and magnitude of the magnetic field at a point midway between the wires. 3.00 A 5.00 A 20.0 cm
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Magnetic Field of a current loop Magnetic field produced by a wire can be enhanced by having the wire in a loop. x1x1 I x2x2 B N loops Current NI 1 loopCurrent I
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[Q] What is the magnetic field at point Q in Fig.? [Q] What is the magnitude and direction of the magnetic field at point P in Fig.?
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[Q] Use the Biot-Savart Law to calculate the magnetic field B at C, the common center of the semicircular area AD and HJ of radii R1=8 cm and R2=4 cm, forming part of the circuit ADJHA carrying current I=10 A, as seen figure.
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Ampere’s Law Consider a circular path surrounding a current, divided in segments l, Ampere showed that the sum of the products of the field by the length of the segment is equal to o times the current. I r B ll
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Consider a case where B is constant and uniform: Then one finds:
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Magnetic Force between two parallel conductors
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l d 1 2 F1F1 B2B2 I1I1 I2I2 Force per unit length
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Definition of the SI unit Ampere If two long, parallel wires 1 m apart carry the same current, and the magnetic force per unit length on each wire is 2x10-7 N/m, then the current is defined to be 1 A. Used to define the SI unit of current called Ampere.
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Example Two wires, each having a weight per units length of 1.0x10 -4 N/m, are strung parallel to one another above the surface of the Earth, one directly above the other. The wires are aligned north-south. When their distance of separation is 0.10 mm what must be the current in each in order for the lower wire to levitate the upper wire. (Assume the two wires carry the same current). l d 1 2 I1I1 I2I2
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l d 1 2 F1F1 B2B2 I1I1 I2I2 mg/l
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Magnetic Field of a solenoid Solenoid magnet consists of a wire coil with multiple loops. It is often called an electromagnet.
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Solenoid Magnet Field lines inside a solenoid magnet are parallel, uniformly spaced and close together. The field inside is uniform and strong. The field outside is non uniform and much weaker. One end of the solenoid acts as a north pole, the other as a south pole. For a long and tightly looped solenoid, the field inside has a value:
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Solenoid Magnet n = N/l : number of (loop) turns per unit length. I : current in the solenoid.
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Example: Consider a solenoid consisting of 100 turns of wire and length of 10.0 cm. Find the magnetic field inside when it carries a current of 0.500 A.
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