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Phys chapter 22 notes

# Phys chapter 22 notes - CHAPTER 22 MAGNETIC FORCES AND...

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CHAPTER - 22 MAGNETIC FORCES AND MAGNETIC FIELDS A stationary charge q experiences a force F in an electric field E according to F = q E. A charge q moving with velocity v experiences a force F in a magnetic field B according to F = q vxB. The magnetic field B is measured in units of Tesla (T) . The direction of F can be determined by first finding the direction of vxB using the right- hand rule (see below). The direction of F is the same as the direction of vxB for a positive charge and opposite to that of vxB for a negative charge. When v and B are expressed in the unit vector ( i,j,k )- notation, the force is calculated algebraically using the rules of vector cross-multiplication ( ixj = k ; jxk = i ; kxi =j , and ixi = jxj = kxk = 0). See solved Example-2 below A trajectory of a charge in a magnetic field will be circular. Make sure you understand how to deal with such a circular motion, see section 22.3 for more details.

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You should also read the sections that deal with: 1. force on a current carrying wire. You should know how the equation describing the force F = I L x B can be used to find the torque on a current-carrying loop in a magnetic field. 2. Biot-Savart law 3. Ampere’s law (learn how to determine the B field for a long wire). 4. Force between two current carrying wires. 5. How to calculate the net B -field produced by several current-carrying conductors.(see solved Example-3 below)
A FEW SOLVED EXAMPLES 1 . A particle of mass, m = 10 -30 kg and charge, q=3.2x 10 -19 C of unknown polarity moving with a velocity of 10 6 m/s enters a region (to the right of line AC as shown in the accompanying diagram) of uniform magnetic field B of magnitude 10.0G . The B -field points out of the page and is perpendicular to the velocity of the charge. After entering the B - field, the charge moves in a circular trajectory as shown [a] Determine the radius of the circular trajectory. qvB =mv 2 /r or r = mv/qB = 10 -30 10 6 /3.2x10 -19 x10 -3 = 3.13x10 -3 m [b] What is the polarity ( + or - ) of the charge ? Ans. The charge is positive [c] How much work was done on the particle by the magnetic field as the particle traveled the half circle ? Ans. Zero. The magnetic force and the displacement are perpendicular to each other. 2 a. A proton of mass M and charge Q = 1.6x10 -19 C is traveling with uniform velocity V = 2x10 5 k m/s under the combined influence of a magnetic field B = 0.6 i + 0.5 j T and a uniform E -field.

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