 # W11D2 Concept Questions Review

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W11D2 Concept Questions Review
Class 18

W07D1 Magnetic Dipoles, Force and Torque on a Dipole, Experiment 2
W07D1 Magnetic Dipoles, Torque and Force on a Dipole, Experiment 2: Magnetic Dipole in a Helmholtz Coil Reading Course Notes: Sections 8.4,8.11.6, Class 18

Concept Question: Magnetic Field Lines
The picture shows the field lines outside a permanent magnet The field lines inside the magnet point: Up Down Left to right Right to left The field inside is zero I don’t know Class 15

Concept Q. Answer: Magnetic Field Lines
Answer: 1. They point up inside the magnet Magnetic field lines are continuous. E field lines begin and end on charges. There are no magnetic charges (monopoles) so B field lines never begin or end Class 15

Concept Question: Parallel Wires
Consider two parallel current carrying wires. With the currents running in the opposite direction, the wires are attracted (opposites attract?) repelled (opposites repel?) pushed another direction not pushed – no net force I don’t know Class 18

Answer: 1. The wires are repelled I1 creates a magnetic field into the page at wire 2. That makes a force on wire 2 to the right. I2 creates a magnetic field into the page at wire 1. That makes a force on wire 1 to the left. Class 18

Concept Question: Dipole in Uniform Magnetic Field
From rest, the coil in a uniform magnetic field above will: rotate wise, not move rotate counterclockwise, not move move to the right, not rotate move to the left, not rotate move in another direction, without rotating both move and rotate neither rotate nor move I don’t know Class 18

Concept Q. Answer: Dipole in Field
Answer: 1. Coil will rotate clockwise (not move) No net force so no center of mass motion. BUT Magnetic dipoles rotate to align with external field (think compass) Class 18

Concept Question: Dipole in Field
The current carrying coil above will feel a net force upwards downwards of zero I don’t know Class 18

Concept Q. Answer: Dipole in Field
Answer: 2. Feels downward force. The forces shown produce a net downward force Class 18

Concept Question: Dipole in Helmholtz
Week 08, Day 2 Week 08, Day 2Week 08, Day 2 Concept Question: Dipole in Helmholtz A dipole pointing along the positive x-direction and located at the center of a Helmholtz coil will feel: a force but not a torque. a torque but not a force. both a torque and a force. neither force nor torque. Class 19 Class 19Class 19 11

Concept Q. Answer: Dipole in Helmholtz
Week 08, Day 2 Week 08, Day 2Week 08, Day 2 Concept Q. Answer: Dipole in Helmholtz Answer: 2. a torque but not a force. The Helmholtz coil makes a UNIFORM FIELD. Dipole feels only torque (need gradient for force). Class 19 Class 19Class 19 12

Concept Question: Dipole in Anti-Helmholtz Coil
Week 08, Day 2 Week 08, Day 2Week 08, Day 2 Concept Question: Dipole in Anti-Helmholtz Coil A dipole pointing along the positive z-direction and located at the center of an anti- Helmholtz coil will feel: a force but not a torque. a torque but not a force. both a torque and a force. neither force nor torque. Class 19 Class 19Class 19 13

Concept Q. Answer: Dipole in Anti-Helmholtz Coil
Week 08, Day 2 Week 08, Day 2Week 08, Day 2 Concept Q. Answer: Dipole in Anti-Helmholtz Coil Answer: 1. A force because there is a non-gradient of the magnetic field but no torque because the magnetic field at the center is zero. Class 19 Class 19Class 19 14

W09D1: Sources of Magnetic Fields: Ampere’s Law
Class 18

Ampere’s Law Class 20

Concept Question: Ampere’s Law
Integrating B around the loop shown gives us: a positive number a negative number zero Class 20

Week 07, Day 2 C.Q. Answer: Ampere’s Law Answer: 3. Total enclosed current is zero, so Class 16

Concept Question: Ampere’s Law
Integrating B around the loop in the clockwise direction shown gives us: a positive number a negative number zero Class 20

Week 07, Day 2 C.Q. Answer: Ampere’s Law Answer: 2. Net enclosed current is out of the page, so field is counter-clockwise (opposite to circulation direction) Class 16

Concept Question: Loop in Uniform Field
Week 09, Day 1 Concept Question: Loop in Uniform Field While a rectangular wire loop is pulled upward though a uniform magnetic field B field penetrating its bottom half, as shown, there is a current in the loop. no current in the loop. I do not understand the concepts of current and magnetic field. I understand the concepts of current and magnetic field but am not sure of the answer. 22 Class 21 22

Concept Q. Ans.: Loop in Uniform Field
Week 09, Day 1 Concept Q. Ans.: Loop in Uniform Field Answer: 1. The motion changes the magnetic flux through the loop. The magnetic flux is decreasing in time as more of the loop enters a region of zero magnetic field. According to Faraday’s Law there is an induced current through the loop. 23 Class 21 23

Concept Q.: Loop in Uniform Field
Week 09, Day 1 Concept Q.: Loop in Uniform Field While a rectangular wire loop is pulled sideways though a uniform magnetic field B field penetrating its bottom half, as shown, there is a current in the loop. no current in the loop. I do not understand the concepts of current and magnetic field. I understand the concepts of current and magnetic field but am not sure of the answer. 24 Class 21 24

Concept Q. Ans.: Loop in Uniform Field
Week 09, Day 1 Concept Q. Ans.: Loop in Uniform Field Answer: 2. The motion does not change the magnetic flux through the loop. The magnetic flux is constant in time. According to Faraday’s Law there is no induced current through the loop. 25 Class 21 25

Concept Question: Loop
The magnetic field through a wire loop is pointed upwards and increasing with time. The induced current in the coil is Clockwise as seen from the top Counterclockwise Class 22

Answer: 1. Induced current is clockwise This produces an “induced” B field pointing down over the area of the loop. The “induced” B field opposes the increasing flux through the loop – Lenz’s Law Class 22

Concept Question: Moving Loop
Week 9, Day 2 Concept Question: Moving Loop A circuit in the form of a rectangular piece of wire is pulled away from a long wire carrying current I in the direction shown in the sketch. The induced current in the rectangular circuit is Clockwise Counterclockwise Neither, the current is zero 28 Class 22 28

Week 9, Day 2 Concept Q. Answer: Moving Loop Answer: 1. Induced current is clockwise B due to I is into page; the flux through the circuit due to that field decreases as the circuit moves away. So the induced current is clockwise (to make a B into the page) Note: Iind dl x B force is left on the left segment and right on the right, but the force on the left is bigger. So the net force on the rectangular circuit is to the left, again trying to keep the flux from decreasing by slowing the circuit’s motion 29 Class 22 29

A coil moves up from underneath a magnet with its north pole pointing upward. The current in the coil and the force on the coil: Current clockwise; force up Current counterclockwise; force up Current clockwise; force down Current counterclockwise; force down Class 22

Answer: 3. Current is clockwise; force is down The clockwise current creates a self-field downward, trying to offset the increase of magnetic flux through the coil as it moves upward into stronger fields (Lenz’s Law). The I dl x B force on the coil is a force which is trying to keep the flux through the coil from increasing by slowing it down (Lenz’s Law again). Class 22

W10D1: Inductance and Magnetic Field Energy
Class 18

Concept Question: Solenoid
A very long solenoid consisting of N turns has radius R and length d, (d>>R).  Suppose the number of turns is halved keeping all the other parameters fixed. The self inductance remains the same. doubles. is halved. is four times as large. is four times as small. None of the above. Class 23

Concept Q. Ans.: Solenoid
Solution 5. The self-induction of the solenoid is equal to the total flux through the object which is the product of the number of turns time the flux through each turn. The flux through each turn is proportional to the magnitude of magnetic field which is proportional to the number of turns per unit length or hence proportional to the number of turns. Hence the self-induction of the solenoid is proportional to the square of the number of turns. If the number of turns is halved keeping all the other parameters fixed then he self inductance is four times as small. Class 23