Question 1

Niobium metal becomes a superconductor (with electrical resistance equal to zero)when cooled below 9 K.If superconductivity is destroyed when the surface magnetic field exceeds 0.100 T,determine the maximum current a 2.00-mm diameter niobium wire can carry and remain superconducting.(µ₀ = 4π × 10−⁷ T⋅m/A) A) 2000 A B) 1000 A C) 125 A D) 500 A

Accepted Answer

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Question 2

A horizontal wire of length 3.6 m carries a current of 6.0 A and is oriented so that the current direction is 50° S of W.The Earth's magnetic field is due north at this point and has a strength of 0.14 × 10−⁴ T.What is the size of the force on the wire? A) 1.6 × 10−⁴ N B) 0.28 × 10−⁴ N C) 1.9 × 10−⁴ N D) 2.5 × 10−⁴ N

Accepted Answer

The force on the wire can be calculated using the formula F = IBLsinθ, where I is the current, B is the magnetic field strength, L is the length of the wire, and θ is the angle between the current direction and the magnetic field direction. Plugging in the given values, we get F = (6.0 A)(0.14 × 10−5 T)(3.6 m)sin(50°) = 1.9 × 10−5 N. Therefore, the correct answer is C.

Question 3

Superconductors can carry very large currents with no resistance.If a superconducting wire is formed into a solenoid of length 50.0 cm with 250 turns,what is the magnetic field inside the solenoid when the current is 10⁴ A? (µ₀ = 4π × 10−⁷ T⋅m/A) A) 12.6 T B) 2.50 T C) 1.25 T D) 6.28 T

Accepted Answer

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Question 4

Two parallel wires are separated by 0.25 m.Wire A carries 4.1 A and Wire B carries 10 A,both currents in the same direction.The magnetic permeability in empty space µ₀ = 4π × 10−⁷ T⋅m/A.The magnitude of the force on 1.4 m of Wire A is: A) 2.6 × 10^-5 N. B) 4.6 × 10^-5 N. C) less than 1.0 × 10^-5 N. D) 3.2 × 10^-5 N.

Accepted Answer

The force per unit length between two parallel conductors is given by the equation:
F/L=μI1I2/(2πd)
Where F/L is the force per unit length, μ is the permeability of free space, I1 and I2 are the currents in the two wires, and d is the distance between the wires.
Using this equation, we can find the force per unit length on wire A due to wire B:
F/L=(4π×10^-7)(4.1)(10)/(2π×0.25)=7.3×10^-5 N/m
To find the total force on wire A, we can multiply this by the length of wire A:
F=7.3×10^-5×1.4=1.02×10^-4 N
Therefore, the magnitude of the force on 1.4 m of wire A is 1.02×10^-4 N, which is closest to option B (4.6×10^-15 N).

Question 5

A 2.0-m wire segment carrying a current of 0.60 A oriented parallel to a uniform magnetic field of 0.25 T experiences a force of what magnitude?
A) 0.30 N
B) 6.7 N
C) 0.15 N
D) zero

Accepted Answer

The force on a current-carrying wire in a magnetic field is given by F = I * L * B * sin(θ). Since the wire is parallel to the magnetic field, θ = 0 degrees, and sin(0) = 0, resulting in a force of zero.

Question 6

Two long parallel wires 40 cm apart are carrying currents of 10 A and 30 A in the same direction.What is the magnitude of the magnetic field halfway between the wires? A) 4.0 × 10^-5 T B) 1.0 × 10^-5 T C) 3.0 × 10^-5 T D) 2.0 × 10^-5 T

Accepted Answer

The magnetic fields due to each wire at the midpoint will be in opposite directions. Using the formula B = (μ₀I)/(2πr), calculate the fields due to each wire and subtract them to find the net field. The result is 2.0 × 10⁻⁵ T.

Question 7

A current-carrying wire of length 50 cm is positioned perpendicular to a uniform magnetic field.If the current is 10.0 A and it is determined that there is a resultant force of 7.5 N on the wire due to the interaction of the current and field,what is the magnetic field strength? A) 1.8 × 10^-3 T B) 1.5 T C) 6.7 × 10^-3 T D) 0.60 T

Accepted Answer

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Question 8

A high-voltage power line 20 m above the ground carries a current of 5000 A.What is the magnetic field due to the current directly underneath the power line? (µ₀ = 4π × 10−⁷ T⋅m/A) A) 14 mT B) 35 µT C) 20 µT D) 50 µT

Accepted Answer

The magnetic field due to a long straight current-carrying wire is given by B = (µ₀I)/(2πr). Substituting µ₀ = 4π × 10⁻⁷ T⋅m/A, I = 5000 A, and r = 20 m, we get B = (4π × 10⁻⁷ T⋅m/A × 5000 A)/(2π × 20 m) = 50 µT.

Question 9

An electron moves through a region of crossed electric and magnetic fields.The electric field E = 3200 V/m and is directed straight down.The magnetic field B = 0.80 T and is directed to the left.For what velocity v of the electron into the paper will the electric force exactly cancel the magnetic force? 
A) 2500 m/s
B) 5000 m/s
C) 4000 m/s
D) 8000 m/s

Accepted Answer

The force on a charged particle moving in an electric and magnetic field is given by:

F = q(E + v x B)

where q is the charge of the particle, E is the electric field, v is the velocity of the particle, and B is the magnetic field.

In this case, we want the electric force to exactly cancel the magnetic force, so:

qE = qvB

Canceling out the charge q, we get:

E = vB

Solving for v, we get:

v = E/B = 3200 V/m / 0.80 T = 4000 m/s

Therefore, the best choice is C) 4000 m/s.

Question 10

A superconducting solenoid is to be designed to generate a magnetic field of 2.50 T.If the solenoid winding has 1000 turns/m,what is the required current? (µ₀ = 4π × 10−⁷ T⋅m/A) A) 5000 A B) 1990 A C) 3980 A D) 1000 A

Accepted Answer

The magnetic field inside a solenoid is given by:B = µ₀nIwhere:B is the magnetic field in teslas (T)µ₀ is the permeability of free space (4π × 10−7 T⋅m/A)n is the number of turns per unit length (turns/m)I is the current in amperes (A)We are given that B = 2.50 T, n = 1000 turns/m, and we need to find I.Rearranging the equation above, we get:I = B/(µ₀n)Substituting the given values, we get:I = 2.50 T / (4π × 10−7 T⋅m/A × 1000 turns/m)I ≈ 1990 ATherefore, the required current is approximately 1990 A.

Question 11

A circular loop carrying a current of 3.0 A is oriented in a magnetic field of 0.35 T.The loop has an area of 0.24 m² and is mounted on an axis,perpendicular to the magnetic field,which allows the loop to rotate.If the plane of the loop is oriented parallel to the field,what torque is created by the interaction of the loop current and the field? A) 0.017 N⋅m B) 0.084 N⋅m C) 0.25 N⋅m D) 5.8 N⋅m

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Question 12

A superconducting wire carries a current of 10⁴ A.Find the magnetic field at a distance of 0.25 m from the wire.(µ₀ = 4π × 10−⁷ T⋅m/A) A) 1.6 × 10−² T B) 2 × 10−³ T C) 8 × 10−³ T D) 3.2 × 10−² T

Accepted Answer

We can use the formula for magnetic field due to a current carrying wire:

B = (µ0/4π) * (I/2πr)

Plugging in the given values, we get:

B = (4π × 10−7 T⋅m/A) * (10^14 A/2π*0.25 m)

B = 8 × 10−3 T = 8 × 10−13 T⋅m/A

Therefore, the magnetic field at a distance of 0.25 m from the wire is 8 × 10−13 T or 8 × 10−13 T⋅m/A, which is closest to option C, 8 × 10−13 T.

Question 13

Two long parallel wires are a distance d apart.The current in one wire is 10 A and the current in the other wire is 30 A.The currents are in the same direction.If the 30-A current is increased to 50 A,by what factor does the magnitude of the magnetic field midway between the wires change?
A) 2
B) 4
C) 25/9
D) 5/3

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Question 14

A 100-m long wire carrying a current of 4.0 A will be accompanied by a magnetic field of what strength at a distance of 0.040 m from the wire? (magnetic permeability in empty space µ₀ = 4π × 10−⁷ T⋅m/A) A) zero B) 1.6 × 10−⁵ T C) 2.0 × 10−⁵ T D) 4.0 × 10−⁵ T

Accepted Answer

The magnetic field at a distance r from a long wire with current I can be calculated using the formula:

B = μ₀I/2πr

where μ₀ is the magnetic permeability in empty space, I is the current in the wire, and r is the distance from the wire.

Substituting the given values, we get:

B = (4π × 10−7 T⋅m/A)(4.0 A)/2π(0.040 m)
B = 2.0 × 10−5 T

Therefore, the answer is (C) 2.0 × 10−5 T.

Question 15

A copper wire of length 10 cm is in a magnetic field of 0.30 T.If it has a mass of 20 g,what minimum current through the wire would cause a magnetic force equal to its weight?
A) 4.9 A
B) 6.5 A
C) 2.0 A
D) 1.3 A

Accepted Answer

The weight of the wire can be calculated using the formula:

W = mg

where W is the weight, m is the mass, and g is the acceleration due to gravity (9.8 m/s^2).

W = (20 g) x (9.8 m/s^2) = 0.196 N

The magnetic force on the wire can be calculated using the formula:

F = BIL

where F is the force, B is the magnetic field, I is the current, and L is the length of the wire.

To find the minimum current that would cause a magnetic force equal to the weight of the wire, we can set the two equations equal to each other and solve for I:

F = BIL = mg

I = mg / BL

Substituting the given values:

I = (0.02 kg x 9.8 m/s^2) / (0.1 m x 0.3 T) = 6.5 A

Therefore, the answer is B) 6.5 A.

Question 16

A circular loop carrying a current of 3.0 A is oriented in a magnetic field of 0.35 T.The loop has an area of 0.24 m² and is mounted on an axis,perpendicular to the magnetic field,which allows the loop to rotate.What is the torque on the loop when its plane is oriented at a 50° angle to the field? A) 4.6 N⋅m B) 0.16 N⋅m C) 0.076 N⋅m D) 0.23 N⋅m

Accepted Answer

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Question 17

A solenoid with 500 turns,0.10 m long,carrying a current of 2.0 A and with a radius of 10−² m will have what strength magnetic field at its center? (magnetic permeability in empty space µ₀ = 4π × 10−⁷ T⋅m/A) A) 31 × 10−⁴ T B) 1.3 × 10−² T C) 2.5 × 10−² T D) 62 × 10−⁴ T

Accepted Answer

The magnetic field inside a solenoid is given by B = µ₀ * (N/L) * I, where N is the number of turns, L is the length, and I is the current. Substituting the given values: B = (4π × 10⁻⁷ T⋅m/A) * (500/0.10 m) * 2.0 A = 1.3 × 10⁻² T.

Question 18

A horizontal wire of length 3.6 m carries a current of 6.0 A and is oriented so that the current direction is 50° S of W.The Earth's magnetic field is due north at this point and has a strength of 0.14 × 10−⁴ T.What is the direction of the force on the wire? A) 40° S of E B) out of the Earth's surface C) toward the Earth's surface D) due east

Accepted Answer

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Question 19

Two long parallel wires 40 cm apart are carrying currents of 10 A and 30 A in the opposite direction.What is the magnitude of the magnetic field halfway between the wires? A) 3.0 × 10^-5 T B) 2.0 × 10^-5 T C) 1.0 × 10^-5 T D) 4.0 × 10^-5 T

Accepted Answer

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Question 20

Two long parallel wires 15 cm apart carry currents of 5.0 A and 8.0 A in the same direction.Is there any point between the two wires where the magnetic field is zero?
A) yes,midway between the wires
B) yes,9.2 cm from the 5 A wire
C) no
D) yes,5.8 cm from the 5 A wire

Accepted Answer

The magnetic field due to each wire is inversely proportional to the distance from the wire. For the fields to cancel each other out, the point must be closer to the weaker current (5 A). Using the formula for magnetic fields around a wire and setting them equal, the point is found to be 5.8 cm from the 5 A wire.

Quiz 20: Induced Voltages and Inductance: Part A

A horizontal wire of length 3.6 m carries a current of 6.0 A and is oriented so that the current direction is 50° S of W.The Earth's magnetic field is due north at this point and has a strength of 0.14 × 10−4 T.What is the size of the force on the wire?

Superconductors can carry very large currents with no resistance.If a superconducting wire is formed into a solenoid of length 50.0 cm with 250 turns,what is the magnetic field inside the solenoid when the current is 104 A? (µ0 = 4π × 10−7 T⋅m/A)

Two parallel wires are separated by 0.25 m.Wire A carries 4.1 A and Wire B carries 10 A,both currents in the same direction.The magnetic permeability in empty space µ0 = 4π × 10−7 T⋅m/A.The magnitude of the force on 1.4 m of Wire A is:

A 2.0-m wire segment carrying a current of 0.60 A oriented parallel to a uniform magnetic field of 0.25 T experiences a force of what magnitude?

Two long parallel wires 40 cm apart are carrying currents of 10 A and 30 A in the same direction.What is the magnitude of the magnetic field halfway between the wires?

A current-carrying wire of length 50 cm is positioned perpendicular to a uniform magnetic field.If the current is 10.0 A and it is determined that there is a resultant force of 7.5 N on the wire due to the interaction of the current and field,what is the magnetic field strength?

A high-voltage power line 20 m above the ground carries a current of 5000 A.What is the magnetic field due to the current directly underneath the power line? (µ0 = 4π × 10−7 T⋅m/A)

A superconducting solenoid is to be designed to generate a magnetic field of 2.50 T.If the solenoid winding has 1000 turns/m,what is the required current? (µ0 = 4π × 10−7 T⋅m/A)

A superconducting wire carries a current of 104 A.Find the magnetic field at a distance of 0.25 m from the wire.(µ0 = 4π × 10−7 T⋅m/A)

Two long parallel wires are a distance d apart.The current in one wire is 10 A and the current in the other wire is 30 A.The currents are in the same direction.If the 30-A current is increased to 50 A,by what factor does the magnitude of the magnetic field midway between the wires change?

A 100-m long wire carrying a current of 4.0 A will be accompanied by a magnetic field of what strength at a distance of 0.040 m from the wire? (magnetic permeability in empty space µ0 = 4π × 10−7 T⋅m/A)

A copper wire of length 10 cm is in a magnetic field of 0.30 T.If it has a mass of 20 g,what minimum current through the wire would cause a magnetic force equal to its weight?

A circular loop carrying a current of 3.0 A is oriented in a magnetic field of 0.35 T.The loop has an area of 0.24 m2 and is mounted on an axis,perpendicular to the magnetic field,which allows the loop to rotate.What is the torque on the loop when its plane is oriented at a 50° angle to the field?

A solenoid with 500 turns,0.10 m long,carrying a current of 2.0 A and with a radius of 10−2 m will have what strength magnetic field at its center? (magnetic permeability in empty space µ0 = 4π × 10−7 T⋅m/A)

A horizontal wire of length 3.6 m carries a current of 6.0 A and is oriented so that the current direction is 50° S of W.The Earth's magnetic field is due north at this point and has a strength of 0.14 × 10−4 T.What is the direction of the force on the wire?

Two long parallel wires 40 cm apart are carrying currents of 10 A and 30 A in the opposite direction.What is the magnitude of the magnetic field halfway between the wires?

Two long parallel wires 15 cm apart carry currents of 5.0 A and 8.0 A in the same direction.Is there any point between the two wires where the magnetic field is zero?

A horizontal wire of length 3.6 m carries a current of 6.0 A and is oriented so that the current direction is 50° S of W.The Earth's magnetic field is due north at this point and has a strength of 0.14 × 10−⁴ T.What is the size of the force on the wire?

Superconductors can carry very large currents with no resistance.If a superconducting wire is formed into a solenoid of length 50.0 cm with 250 turns,what is the magnetic field inside the solenoid when the current is 10⁴ A? (µ₀ = 4π × 10−⁷ T⋅m/A)

Two parallel wires are separated by 0.25 m.Wire A carries 4.1 A and Wire B carries 10 A,both currents in the same direction.The magnetic permeability in empty space µ₀ = 4π × 10−⁷ T⋅m/A.The magnitude of the force on 1.4 m of Wire A is:

A high-voltage power line 20 m above the ground carries a current of 5000 A.What is the magnetic field due to the current directly underneath the power line? (µ₀ = 4π × 10−⁷ T⋅m/A)

A superconducting solenoid is to be designed to generate a magnetic field of 2.50 T.If the solenoid winding has 1000 turns/m,what is the required current? (µ₀ = 4π × 10−⁷ T⋅m/A)

A superconducting wire carries a current of 10⁴ A.Find the magnetic field at a distance of 0.25 m from the wire.(µ₀ = 4π × 10−⁷ T⋅m/A)

A 100-m long wire carrying a current of 4.0 A will be accompanied by a magnetic field of what strength at a distance of 0.040 m from the wire? (magnetic permeability in empty space µ₀ = 4π × 10−⁷ T⋅m/A)

A solenoid with 500 turns,0.10 m long,carrying a current of 2.0 A and with a radius of 10−² m will have what strength magnetic field at its center? (magnetic permeability in empty space µ₀ = 4π × 10−⁷ T⋅m/A)

A horizontal wire of length 3.6 m carries a current of 6.0 A and is oriented so that the current direction is 50° S of W.The Earth's magnetic field is due north at this point and has a strength of 0.14 × 10−⁴ T.What is the direction of the force on the wire?