Multiple Choice Practice Problems Slide 3 / 50 Slide 4 / 50 1 A - - PDF document

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Magnetism Practice Problems

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Multiple Choice

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1 A straight wire carries a current down. What is the direction of the magnetic field at the point to the East from the wire? A West B East C North D South E Down

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2 A loop of wire carries a current in counterclockwise direction. What is the direction

• f the magnetic field inside the loop?

A Point to the left B Points to the right C Points out of the page D Points into the page E Curls in a clockwise direction

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3 A current carrying wire is placed horizontally and has a current flow to the right. What is the direction of the magnetic field at point P? A Points to the right B Points to the left C Points to the top of the page D Points to the bottom of the page E Points out of the page

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4 A vertical wire carries a current straight up in a region of the magnetic field directed north. What is the direction of the magnetic force on the current due to the magnetic field? A East B South C North D West E Applied force is zero

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5 A horizontal thin wire has a mass m and length L. The wire carries a constant current I. What must be the direction of the magnetic field in order to cancel the gravitational force? A Left B Right C Down the page D Out of the page E Into the page

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6 A horizontal thin wire has a mass m and length L. The wire carries a constant current I. What must be the magnitude of the magnetic field in order to cancel the gravitational force? A IL/mg B mg/IL C Ig/mL D Im/gL E Zero

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7 An electron enters a uniform magnetic field directed in –Z. What is the direction of the magnetic force on the electron due to the magnetic field? A +X direction B +Y direction C

• X direction

D –Y direction E Applied force is zero

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8 A proton enters a uniform magnetic field directed in +Z. What is the direction of the magnetic force

• n the proton due to the magnetic field?

A +X direction B +Y direction C

• X direction

D –Y direction E Applied force is zero

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9 A horizontal wire carries a current to the east. A proton moves to the west in the region north from the current. What is the direction of the magnetic force on the proton? A West B North C East D South E Applied force is zero

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10 A circular loop of wire carries a constant current in counterclockwise direction. The loop is placed in a uniform magnetic field directed into the page. What is the effect of the magnetic force on the loop? A Rotates with respect to its axis B Rotates with respect to its diameter C Contracts its size D Expends its size E No effect on the loop

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11 A proton enters a uniform magnetic field perpendicular to the field lines. What is the new path of the proton as it passes the field? A A B B C C D D E E

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12 A magnetic field is created by two parallel currents flowing in opposite directions. At which location the magnetic field is greatest in magnitude? A A B B C C D D E E

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13 Two parallel wires carry currents in opposite

• directions. What is the direction on the net

magnetic field at point P? A Left B Right C Top the page D Bottom the page E Into the page

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14 Two parallel wires carry currents in the same

• direction. What is the direction of the magnetic

force on current I2 due to current I1? A Left B Right C Top the page D Bottom the page E Out of the page

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15 Two parallel wires carry currents in opposite

• directions. What is the direction of the magnetic

force on current I1 due to current I2? A Left B Right C Top the page D Bottom the page E Out of the page

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16 An electron enters a uniform electric field perpendicular to the field lines. What must be the direction of the magnetic field in order to cancel the electric force effect? A Left B Right C Top of the page D Into the page E Out of the page

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17 An electron enters a uniform electric field perpendicular to the field lines. What is the magnitude of the magnetic field if the electric effect completely canceled? A Ev B v/E C zero D E/v E eEv

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18 What is the magnitude of the magnetic field at point B produced by a current I if the magnitude of the field at point A is B0? A B0 B 2B0 C 4B0 D B0/2 E B0/4

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19 Two parallel wires carry currents I1 and I2 in the same direction and separated by a distance d. The magnitude of the magnetic force between the wires is F0. What is the force between the wires if each current is doubled and the separation is quadrupled? A 2F0 B 4F0 C F0 D F0/2 E F0/4

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20 An electron with a mass m and charge e enters at a constant speed v a uniform magnetic field B. What is the radius of the curvature of the electron in the field? A mv/eB B eB/mv C me/vB D mB/ev E zero

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Free Response

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• 1. A long horizontal wire carries an electric current I = 50 A. Point P is located

at a distance 2.5 mm above the current.

• a. What is the direction of the magnetic field at point P?
• b. What is the magnitude of the magnetic field at point P?

A thin horizontal rod has a length L = 1 and mass m = 50 g is connected to a circuit. The circuit contains a battery V = 12 V, a resistor R = 0.06 Ω, a switch, and connecting

• wires. The rod is supported in horizontal position by two

light connecting wires.

• c. What is the direction of the electric current in the rod?
• d. On the diagram show all the applied forces on the rod.
• e. What is the tension force in supporting wires?

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• 1. A long horizontal wire carries an electric current I = 50 A. Point P is located

at a distance 2.5 mm above the current.

• a. What is the direction of the magnetic field at point P?

Out of the page.

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• 1. A long horizontal wire carries an electric current I = 50 A. Point P is located

at a distance 2.5 mm above the current.

• b. What is the magnitude of the magnetic field at point P?

B = (μ0/2π)(I/r) B = (2x107 Tm/A)/(50A/0.0025m) B = 0.004T

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1. A thin horizontal rod has a length L = 1 m and mass m = 50 g is connected to a circuit. The circuit contains a battery V = 12 V, a resistor R = 0.06 Ω, a switch, and connecting wires. The rod is supported in horizontal position by two light connecting wires.

• c. What is the direction of the electric current in the rod?

To the right.

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1. A thin horizontal rod has a length L = 1 m and mass m = 50 g is connected to a circuit. The circuit contains a battery V = 12 V, a resistor R = 0.06 Ω, a switch, and connecting wires. The rod is supported in horizontal position by two light connecting wires.

• d. On the diagram show all the applied forces on the rod.

FB mg

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1. A thin horizontal rod has a length L = 1 m and mass m = 50 g is connected to a circuit. The circuit contains a battery V = 12 V, a resistor R = 0.06 Ω, a switch, and connecting wires. The rod is supported in horizontal position by two light connecting wires.

• e. What is the tension force in supporting wires?

ΣF = ma T + T - FB - mg = 0 2T = FB + mg T = (FB + mg)/2 T = (IlB + mg)/2 T = [(50A)(1m)(0.004T)+(0.05kg)(9.8m/s2)/2 T = 0.35N each

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• 2. Charged particle of mass m and charge q is

released from rest in region between two charged plates M and L. After passing the region of the electric field with an accelerating voltage Va the particle enters another region filled with a magnetic field of magnitude B and directed out of the page.

• a. What is the sign of the charge on the particle?
• b. What is the velocity of the particle as it enters the magnetic field?
• c. What is the direction of the magnetic force on the particle?
• d. Describe the path of the particle in the magnetic field.
• e. What is the radius of the curvature of the particle in the magnetic field?
• f. What is the direction and magnitude of the electric field that can be used

to make the path of the particle straight?

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• 2. Charged particle of mass m and charge q is

released from rest in region between two charged plates M and L. After passing the region of the electric field with an accelerating voltage Va the particle enters another region filled with a magnetic field of magnitude B and directed out of the page.

• a. What is the sign of the charge on the particle?

positive

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• 2. Charged particle of mass m and charge q is

released from rest in region between two charged plates M and L. After passing the region of the electric field with an accelerating voltage Va the particle enters another region filled with a magnetic field of magnitude B and directed out of the page.

• b. What is the velocity of the particle as it enters the magnetic field?

E0 + W = EF E0 = EF qV = ½mv2 qV = ½mv2 v2 = 2qV/m v = (2qV/m)1/2

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• 2. Charged particle of mass m and charge q is

released from rest in region between two charged plates M and L. After passing the region of the electric field with an accelerating voltage Va the particle enters another region filled with a magnetic field of magnitude B and directed out of the page.

• c. What is the direction of the magnetic force on the particle?

Down

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• 2. Charged particle of mass m and charge q is

released from rest in region between two charged plates M and L. After passing the region of the electric field with an accelerating voltage Va the particle enters another region filled with a magnetic field of magnitude B and directed out of the page.

• d. Describe the path of the particle in the magnetic field.

Circular, down, clockwise.

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• 2. Charged particle of mass m and charge q is

released from rest in region between two charged plates M and L. After passing the region of the electric field with an accelerating voltage Va the particle enters another region filled with a magnetic field of magnitude B and directed out of the page.

• e. What is the radius of the curvature of the particle in the magnetic field?

ΣF = ma FB = ma qvB = mv2/r qB = mv/r r = mv/qB r = m(2qV/m)1/2/qB r = (2mVa/qB2)1/2

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• 2. Charged particle of mass m and charge q is

released from rest in region between two charged plates M and L. After passing the region of the electric field with an accelerating voltage Va the particle enters another region filled with a magnetic field of magnitude B and directed out of the page.

• f. What is the direction and magnitude of the electric field that can be used

to make the path of the particle straight? ΣF = ma FE - FB = 0 FE = FB qE = qvB E = vB E = B(2qVa/m)1/2

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• 3. An electron is accelerated by an electric field

produced by two parallel plates M and L. When the electron enters a region filled with a magnetic field of magnitude B = 0.5 T its velocity v = 1.6x107 m/s.

• a. What is the direction of the accelerating

electric field between the plates M and L?

• b. What is the accelerating voltage of the

electric field?

• c. What is the direction of the magnetic

field?

• d. What is the radius of the curvature of the

electron in the magnetic field?

• e. What is the direction of the deflecting

electric field required to make the electron’s path straight?

• f. What is the magnitude of the deflecting

electric field?

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• 3. An electron is accelerated by an electric field

produced by two parallel plates M and L. When the electron enters a region filled with a magnetic field of magnitude B = 0.5 T its velocity v = 1.6x107 m/s.

• a. What is the direction of the accelerating

electric field between the plates M and L? Left

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• 3. An electron is accelerated by an electric field

produced by two parallel plates M and L. When the electron enters a region filled with a magnetic field of magnitude B = 0.5 T its velocity v = 1.6x107 m/s.

• b. What is the accelerating voltage of the

electric field? E0 + W = EF E0 = EF qV = ½mv2 qV = ½mv2 V = mv2/2q V = (9.11x10-31 kg)(1.6x107 m/s)/(2)(1.6x10-19 C) V = 729 V

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• 3. An electron is accelerated by an electric field

produced by two parallel plates M and L. When the electron enters a region filled with a magnetic field of magnitude B = 0.5 T its velocity v = 1.6x107 m/s.

• c. What is the direction of the magnetic

field? Into the page.

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• 3. An electron is accelerated by an electric field

produced by two parallel plates M and L. When the electron enters a region filled with a magnetic field of magnitude B = 0.5 T its velocity v = 1.6x107 m/s.

• d. What is the radius of the curvature of the

electron in the magnetic field? ΣF = ma FB = ma qvB = mv2/r qB = mv/r r = mv/qB r = (9.11x10-31 kg)(1.6x10-7 m/s)/(1.6x10-19 C)(0.5 T) r = 1.8x10-18 m

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• 3. An electron is accelerated by an electric field

produced by two parallel plates M and L. When the electron enters a region filled with a magnetic field of magnitude B = 0.5 T its velocity v = 1.6x107 m/s.

• e. What is the direction of the deflecting

electric field required to make the electron’s path straight? Down

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• 3. An electron is accelerated by an electric field

produced by two parallel plates M and L. When the electron enters a region filled with a magnetic field of magnitude B = 0.5 T its velocity v = 1.6x107 m/s.

• f. What is the magnitude of the deflecting

electric field? ΣF = ma FE - FB = 0 FE = FB qE = qvB E = vB E = (1.6x107 m/s)(0.5 T) E = 8x106 m/s

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• 4. In a mass spectrometer a charged particle is

accelerated to a velocity v = 5.9x107 m/s by an electric field and allowed to enter a magnetic field B, where it is deflected in a semi-circular path of radius R = 10 cm. The magnetic field is uniform of magnitude B = 16 T and oriented out

• f the page.
• a. What is the sign of the charge on the particle?
• b. What is the acceleration of the particle in the

magnetic field?

• c. What is the ratio between the charge and mass
• f the particle q/m?
• d. What is the direction of the accelerating

electric field?

• e. What is the accelerating voltage of the

electric field?

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• 4. In a mass spectrometer a charged particle is

accelerated to a velocity v = 5.9x107 m/s by an electric field and allowed to enter a magnetic field B, where it is deflected in a semi-circular path of radius R = 10 cm. The magnetic field is uniform of magnitude B = 16 T and oriented out

• f the page.
• a. What is the sign of the charge on the particle?

positive

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• 4. In a mass spectrometer a charged particle is

accelerated to a velocity v = 5.9x107 m/s by an electric field and allowed to enter a magnetic field B, where it is deflected in a semi-circular path of radius R = 10 cm. The magnetic field is uniform of magnitude B = 16 T and oriented out

• f the page.
• b. What is the acceleration of the particle in the

magnetic field? a = v2 r (5.9x107 m/s)2 0.1 m = 3.5x1016 m/s2 =

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• 4. In a mass spectrometer a charged particle is

accelerated to a velocity v = 5.9x107 m/s by an electric field and allowed to enter a magnetic field B, where it is deflected in a semi-circular path of radius R = 10 cm. The magnetic field is uniform of magnitude B = 16 T and oriented out

• f the page.
• c. What is the ratio between the charge and mass
• f the particle q/m?

qvB = mv2/r q/m = v/Br q/m = 3.7x107 C/kg

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• 4. In a mass spectrometer a charged particle is

accelerated to a velocity v = 5.9x107 m/s by an electric field and allowed to enter a magnetic field B, where it is deflected in a semi-circular path of radius R = 10 cm. The magnetic field is uniform of magnitude B = 16 T and oriented out

• f the page.
• d. What is the direction of the accelerating

electric field? to the right

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• 4. In a mass spectrometer a charged particle is

accelerated to a velocity v = 5.9x107 m/s by an electric field and allowed to enter a magnetic field B, where it is deflected in a semi-circular path of radius R = 10 cm. The magnetic field is uniform of magnitude B = 16 T and oriented out

• f the page.
• e. What is the accelerating voltage of the

electric field? qV = 1/2mv2 V = mv2/2q V = (3.7x107)-1 (5.9x107 m/s)2 2 V = 4.7x107 V

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