QuestionsCh28 - CHAPTER 28 Sources of Magnetic Field...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
CHAPTER 28: Sources of Magnetic Field Responses to Questions 1. Alternating currents will have little effect on the compass needle, due to the rapid change of the direction of the current and of the magnetic field surrounding it. Direct currents will deflect a compass needle. The deflection depends on the magnitude and direction of the current and the distance from the current to the compass. The effect on the compass decreases with increasing distance from the wire. 2. The magnetic field due to a long straight current is proportional to the current strength. The electric field due to a long straight line of electric charge at rest is proportional to the charge per unit length. Both fields are inversely proportional to the distance from the wire or line of charge. The magnetic field lines form concentric circles around the wire; the electric field lines are directed radially outward if the line of charge is positive and radially inward if the line of charge is negative. 3. The magnetic forces exerted on one wire by the other try to align the wires. The net force on either wire is zero, but the net torque is not zero. 4. Yes. Assume the upper wire is fixed in position. Since the currents in the wires are in the same direction, the wires will attract each other. The lower wire will be held in equilibrium if this force of attraction (upward) is equal in magnitude to the weight of the wire (downward). 5. ( a ) The current in the lower wire is opposite in direction to the current in the upper wire. ( b ) The upper wire can be held in equilibrium due to the balance between the magnetic force from the lower wire and the gravitational force. The equilibrium will be stable for small vertical displacements, but not for horizontal displacements. 6. ( a ) Let 21 . II = () 0e n c l 0 1 2 01 2 dI I I I µµ µ ⋅= = + = B l r r ( b ) Let . =− n c l 0 1 2 0 I I = B l r r Inside the cavity since the geometry is cylindrical and no current is enclosed. 0 = B r 8. Construct a closed path similar to that shown in part ( a ) of the figure, such that sides ab and cd are perpendicular to the field lines and sides bc and da lie along the field lines. Unlike part ( a ), the path will not form a rectangle; the sides ab and cd will flare outward so that side bc is longer than side da . Since the field is stronger in the region of da than it is in the region of bc , but da is shorter than bc , the contributions to the integral in Ampère’s law may cancel.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 4

QuestionsCh28 - CHAPTER 28 Sources of Magnetic Field...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online