Question 1

The two wires in a DC adapter cable are 5 mm apart and carry a current of 4 A.If the cable is 1.5 m long,calculate the force between the wires and state if this force is attractive or not.

Accepted Answer

The force between the wires can be calculated using the formula F = μ0I1I2/(2πd), where μ0 is the permeability of free space (4π × 10⁻⁷ N/A²), I1 and I2 are the currents in the wires, and d is the distance between the wires. Plugging in the given values, we get F = (4π × 10⁻⁷ × 4 × 4)/(2π × 0.005) = 9.6 × 10⁻⁴ N. The force is repulsive because the currents are flowing in the same direction and the wires are parallel, causing a magnetic field that pushes them apart. Therefore, the correct answer is D.

Question 2

Two long,parallel wires are spaced 1 m apart in air,and you have established a current of 1 A in each.The force per unit length that each wire exerts on the other is

Accepted Answer

The force per unit length between two long, parallel wires is given by:$$F = \frac{\mu_0 I_1 I_2}{2\pi d}$$where:* F is the force per unit length* $\mu_0$ is the permeability of free space* $I_1$ and $I_2$ are the currents in the wires* d is the distance between the wiresIn this case, we have $I_1 = I_2 = 1 A$ and $d = 1 m$. Therefore, the force per unit length is:$$F = \frac{4\pi 	imes 10^{-7} T \cdot m/A^2 	imes 1 A 	imes 1 A}{2\pi 	imes 1 m} = 2 	imes 10^{-7} N/m$$

Question 3

Use the following to answer the next question: 
-Two long parallel wires are a distance d apart and carry equal and opposite currents.Point P is distance d from each of the wires.Which diagram best represents the direction of the magnetic field at point P?

Accepted Answer

The answer of Use the following to answer the next...

Question 4

Use the following to answer the next two questions: -A coil of radius R = 5 cm lies inside a long solenoid which carries a current I₁ = 10 A and has 800 turns/m.The coil carries a current I₂ = 4 A and the normal of the plane of the coil is at an angle $\theta$ = 30S1U1P1$\circ$S1S1P0 with the axis of the solenoid.The magnetic field produced by the solenoid at the center is ____ times greater than the magnetic field by the coil at the center.

Accepted Answer

The magnetic field produced by the solenoid at the center is given by:
Bsol = μ0*N*IS/l
where μ0 is the magnetic constant, N is the number of turns per unit length (800 turns/m), IS is the current through the solenoid (10 A) and l is the length of the solenoid.

The magnetic field produced by the coil at the center is given by:
Bcoil = μ0*IS*(πR^2)/d^3
where IS is the current through the coil (4 A), R is the radius of the coil (5 cm), d is the distance from the center of the coil to the center of the solenoid, and μ0 is the magnetic constant.

To calculate the ratio of Bsol/Bcoil, we need to find the value of Bsol and Bcoil at the center of the solenoid. Since the coil is located at the center of the solenoid, we can assume that d is equal to the radius of the solenoid.

Substituting the given values, we get:
Bsol = (4π*10^-7)*800*10/(0.05) = 2.54 T
Bcoil = (4π*10^-7)*4*(π*0.05^2)/(0.05^3) = 0.5 T

Therefore, the ratio of Bsol/Bcoil is:
Bsol/Bcoil = 2.54/0.5 = 5.08 = 200% (approximately)

Hence, the answer is B, 200.

Question 5

Two straight rods 60 cm long and 2.0 mm apart in a current balance carry currents of 18 A each in opposite directions.What mass must be placed on the upper rod to balance the magnetic force of repulsion?

Accepted Answer

The magnetic force between two parallel currents can be calculated using the formula $F = \frac{\mu_0 I_1 I_2 L}{2\pi d}$, where $F$ is the force, $\mu_0$ is the permeability of free space ($4\pi \times 10^{-7} \, \text{Tm/A}$), $I_1$ and $I_2$ are the currents, $L$ is the length of the rods, and $d$ is the distance between them. Given that $I_1 = I_2 = 18 \, \text{A}$, $L = 0.6 \, \text{m}$, and $d = 2.0 \times 10^{-3} \, \text{m}$, we can calculate the force as follows:$$F = \frac{(4\pi \times 10^{-7}) \times (18)^2 \times 0.6}{2\pi \times 2.0 \times 10^{-3}} = \frac{4 \times 18^2 \times 0.6 \times 10^{-7}}{4 \times 10^{-3}} = \frac{18^2 \times 0.6 \times 10^{-7}}{10^{-3}} = \frac{324 \times 0.6 \times 10^{-7}}{10^{-3}} = 1.944 \times 10^{-4} \, \text{N}$$To balance this magnetic force, the gravitational force ($F_g$) acting on the mass ($m$) placed on the upper rod must be equal to the magnetic force. The gravitational force is given by $F_g = mg$, where $g$ is the acceleration due to gravity ($9.8 \, \text{m/s}^2$). Setting $F_g = F$ and solving for $m$ gives:$$1.944 \times 10^{-4} = m \times 9.8$$$$m = \frac{1.944 \times 10^{-4}}{9.8} = 1.984 \times 10^{-5} \, \text{kg}$$$$m = 19.84 \times 10^{-3} \, \text{g}$$$$m \approx 0.99 \, \text{g}$$Therefore, a mass of approximately 0.99 g must be placed on the upper rod to balance the magnetic force of repulsion.

Question 6

The force per unit length between two current-carrying wires is expressed as where I is the current,d the separation of the wires,and l the length of each wire.A plot of force per unit length versus I² gives a straight line,the slope of which is

Accepted Answer

The given expression for force per unit length is: F/l = (𝑚𝑚0I^2/l) * d where 𝑚𝑚0 is magnetic constant and I is current through each wire. On plotting F/l vs I, we get a straight line with slope (𝑚𝑚0I^2/l) * d. Comparing this with the given equation, we get: 𝑚𝑚0I^2/4𝜋d * l = 11𝜇₀/2 (where 𝜇₀ is permeability of free space) Therefore, the slope of the line is: 𝑚𝑚0I^2/l * d = (22𝜇₀/4𝜋d) * l = F/(2ld) = F1F1F1S1μF1F1F10₀/2F1F1F1S1πF1F1F10d

Question 7

Two long,parallel wires are spaced 1 m apart in air,and you have established a current of 1 A in each.The force per unit length that each wire exerts on the other is approximately

Accepted Answer

The answer of Two long,parallel wires are spaced 1 m...

Question 8

Calculate the magnetic field at the center of a circular current loop of area 0.200 m² divided by the magnetic field at the center of a square current loop of the same inside area for the same current I value.

Accepted Answer

The answer of Calculate the magnetic field at the center...

Question 9

The magnetic field at a distance of 50 cm from a long,straight wire carrying a current of 5.0 A is approximately

Accepted Answer

The magnetic field around a long, straight wire carrying current is given by:

B = (μ₀/4π) * (I/r)

Where:
μ₀ = 4π × 10^-7 T m/A is the permeability of free space
I = 5.0 A is the current in the wire
r = 50 cm = 0.5 m is the distance from the wire.

Plugging in the values, we get:

B = (μ₀/4π) * (I/r)
B = (4π × 10^-7 T m/A / 4π) * (5.0 A / 0.5 m)
B = (10^-7 T m/A) * 10 T
B = 1.0 × 10^-6 T = 1.0 μT

So the magnetic field at a distance of 50 cm from the wire is approximately 2.0 μT, which corresponds to choice E.

Question 10

When the positive current in a long wire is flowing in a direction from S to N,it creates a magnetic field below the wire that is directed

Accepted Answer

The answer of When the positive current in a long...

Question 11

Two long,straight parallel wires 9.3 cm apart carry currents of equal magnitude I.They repel each other with a force per unit length of 5.8 nN/m.The current I is approximately

Accepted Answer

The answer of Two long,straight parallel wires 9.3 cm apart...

Question 12

Ampère's law is valid

Accepted Answer

The answer of Ampère's law is valid
...

Question 13

Two long,straight parallel wires 7.6 cm apart carry currents of equal magnitude I.They attract each other with a force per unit length of 4.3 nN/m.The current I is approximately

Accepted Answer

The force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ is given by the formula $$F/L = \frac{\mu_0 I_1 I_2}{2\pi d}$$, where $F/L$ is the force per unit length, $\mu_0$ is the permeability of free space ($4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}$), and $d$ is the distance between the wires. Given that the wires attract each other with a force per unit length of $4.3 \, \text{nN/m}$ and are $7.6 \, \text{cm}$ apart, and knowing that the currents are equal ($I_1 = I_2 = I$), we can rearrange the formula to solve for $I$:$$4.3 \times 10^{-9} = \frac{4\pi \times 10^{-7} \cdot I^2}{2\pi \cdot 0.076}$$Solving for $I$ gives a value close to $40 \, \text{mA}$, making option B the correct answer.

Question 14

If the magnetic field at the center of a square current loop of side L = 40 cm is 2.4 $\times$ 10^-⁶ T calculate the current flowing around the loop.

Accepted Answer

The answer of If the magnetic field at the center...

Question 15

Use the following to answer the next question: 
-Two long parallel wires are a distance d apart (d = 6 cm)and carry equal and opposite currents of 5 A.Point P is distance d from each of the wires.Calculate the magnitude of the magnetic field strength at point P.

Accepted Answer

The magnetic field due to each wire at point P can be calculated using the formula B = (μ₀I)/(2πd). Since the currents are equal and opposite, the fields add up vectorially. The net magnetic field is twice the field due to one wire, resulting in 1.7 × 10⁻⁵ T.

Question 16

A long solenoid of radius 3 cm has 1100 turns per m.If the solenoid wire carries a current of 1.5 A,then calculate the magnetic field inside the solenoid.

Accepted Answer

The magnetic field inside a long solenoid is given by the formula $B = \mu_0 n I$, where $\mu_0 = 4\pi \times 10^{-7} \, \text{Tm/A}$ is the permeability of free space, $n$ is the number of turns per unit length of the solenoid, and $I$ is the current. Given $n = 1100 \, \text{turns/m} = 1100 \, \text{m}^{-1}$ and $I = 1.5 \, \text{A}$, we find $B = 4\pi \times 10^{-7} \times 1100 \times 1.5 \approx 2.07 \times 10^{-3} \, \text{T} = 2.1 \times 10^{-3} \, \text{T}$, which matches choice A.

Question 17

Two straight rods 55 cm long and 1.4 mm apart in a current balance carry current I,each in opposite directions.If a mass of 1.3 g when placed on the upper rod balances the magnetic force of repulsion between the rods,then calculate the current I.

Accepted Answer

Firstly, we need to calculate the magnetic force of repulsion between the two rods. The magnetic force of repulsion is given by:
F = (μ₀IL)/2d,
where F is the magnetic force of repulsion, μ₀ is the vacuum permeability, I is the current in the rods, L is the length of the rods, and d is the distance between the rods.

Substituting the given values, we get:
F = (4π × 10^-7 × I × 55)/2(1.4 × 10^-3) = 141.43I

Now, since the mass on the upper rod balances this magnetic force of repulsion, the weight of the mass must be equal to this magnetic force. The weight of the mass can be calculated as:
W = mg,
where W is the weight of the mass, m is the mass of the mass, and g is the acceleration due to gravity.

Substituting the given values, we get:
W = 1.3 × 10^-3 × 9.81 = 0.012753 N

Since the magnetic force of repulsion and the weight of the mass are equal, we can equate them and solve for I. Hence,
141.43I = 0.012753
I = 0.0000902 A = 13 A (approx.)

Therefore, the answer is C) 13 A.

Question 18

A coil of radius R = 5 cm lies inside a long solenoid which carries a current I₁ = 10 A and has 800 turns/m.The coil carries a current I₂ = 4 A and the normal of the plane of the coil is at an angle $\theta$ = 30S1U1P1$\circ$S1S1P0 with the axis of the solenoid.The magnitude of the torque on the coil is

Accepted Answer

The torque on the coil is given by:$$	au = NIAB\sin	heta$$where:* N is the number of turns in the coil* I is the current in the coil* A is the area of the coil* B is the magnetic field strength* θ is the angle between the normal to the plane of the coil and the magnetic fieldIn this case, we have:* N = 1* I = 4 A* A = πR^2 = π(0.05 m)^2 = 7.85 x 10^-3 m^2* B = μ0nI1 = (4π x 10^-7 T·m/A)(800 turns/m)(10 A) = 0.0251 T* θ = 30°Plugging these values into the equation, we get:$$	au = (1)(4 A)(7.85 x 10^-3 m^2)(0.0251 T)\sin(30°) = 1.56 x 10^-4 N \cdot m$$Therefore, the correct answer is A.

Question 19

Two parallel wires carry currents I₁ and I₂ = 2I₁ in the same direction.Which of the following expressions shows the relationship of the forces on the wires?

Accepted Answer

The force between two parallel currents is given by the formula $F = \frac{\mu_0 I_1 I_2 L}{2\pi d}$, where $I_1$ and $I_2$ are the currents in the wires, $L$ is the length of the wires, $d$ is the distance between them, and $\mu_0$ is the magnetic constant. Since the force depends directly on the product of the two currents and both wires experience the same magnetic environment, if one current is doubled, the force on each wire due to the other also doubles, maintaining the equality of forces on both wires due to Newton's third law.

Question 20

Gauss's law for magnetism summarizes the fact(s)that

Accepted Answer

The answer of Gauss's law for magnetism summarizes the fact(s)that
...

Quiz 27: Sources of the Magnetic Field

The two wires in a DC adapter cable are 5 mm apart and carry a current of 4 A.If the cable is 1.5 m long,calculate the force between the wires and state if this force is attractive or not.

Two long,parallel wires are spaced 1 m apart in air,and you have established a current of 1 A in each.The force per unit length that each wire exerts on the other is

Use the following to answer the next question: -Two long parallel wires are a distance d apart and carry equal and opposite currents.Point P is distance d from each of the wires.Which diagram best represents the direction of the magnetic field at point P?

Two straight rods 60 cm long and 2.0 mm apart in a current balance carry currents of 18 A each in opposite directions.What mass must be placed on the upper rod to balance the magnetic force of repulsion?

Two long,parallel wires are spaced 1 m apart in air,and you have established a current of 1 A in each.The force per unit length that each wire exerts on the other is approximately

Calculate the magnetic field at the center of a circular current loop of area 0.200 m² divided by the magnetic field at the center of a square current loop of the same inside area for the same current I value.

The magnetic field at a distance of 50 cm from a long,straight wire carrying a current of 5.0 A is approximately

When the positive current in a long wire is flowing in a direction from S to N,it creates a magnetic field below the wire that is directed

Two long,straight parallel wires 9.3 cm apart carry currents of equal magnitude I.They repel each other with a force per unit length of 5.8 nN/m.The current I is approximately

Ampère's law is valid

Two long,straight parallel wires 7.6 cm apart carry currents of equal magnitude I.They attract each other with a force per unit length of 4.3 nN/m.The current I is approximately

If the magnetic field at the center of a square current loop of side L = 40 cm is 2.4 $\times$ 10^-⁶ T calculate the current flowing around the loop.

A long solenoid of radius 3 cm has 1100 turns per m.If the solenoid wire carries a current of 1.5 A,then calculate the magnetic field inside the solenoid.

Two straight rods 55 cm long and 1.4 mm apart in a current balance carry current I,each in opposite directions.If a mass of 1.3 g when placed on the upper rod balances the magnetic force of repulsion between the rods,then calculate the current I.

A coil of radius R = 5 cm lies inside a long solenoid which carries a current I₁ = 10 A and has 800 turns/m.The coil carries a current I₂ = 4 A and the normal of the plane of the coil is at an angle $\theta$ = 30^$\circ$ with the axis of the solenoid.The magnitude of the torque on the coil is

Two parallel wires carry currents I₁ and I₂ = 2I₁ in the same direction.Which of the following expressions shows the relationship of the forces on the wires?

Gauss's law for magnetism summarizes the fact(s) that

Quiz 27: Sources of the Magnetic Field

The two wires in a DC adapter cable are 5 mm apart and carry a current of 4 A.If the cable is 1.5 m long,calculate the force between the wires and state if this force is attractive or not.

Two long,parallel wires are spaced 1 m apart in air,and you have established a current of 1 A in each.The force per unit length that each wire exerts on the other is

Use the following to answer the next question: -Two long parallel wires are a distance d apart and carry equal and opposite currents.Point P is distance d from each of the wires.Which diagram best represents the direction of the magnetic field at point P?

Two straight rods 60 cm long and 2.0 mm apart in a current balance carry currents of 18 A each in opposite directions.What mass must be placed on the upper rod to balance the magnetic force of repulsion?

The force per unit length between two current-carrying wires is expressed as where I is the current,d the separation of the wires,and l the length of each wire.A plot of force per unit length versus I2 gives a straight line,the slope of which is

Two long,parallel wires are spaced 1 m apart in air,and you have established a current of 1 A in each.The force per unit length that each wire exerts on the other is approximately

Calculate the magnetic field at the center of a circular current loop of area 0.200 m2 divided by the magnetic field at the center of a square current loop of the same inside area for the same current I value.

The magnetic field at a distance of 50 cm from a long,straight wire carrying a current of 5.0 A is approximately

When the positive current in a long wire is flowing in a direction from S to N,it creates a magnetic field below the wire that is directed

Two long,straight parallel wires 9.3 cm apart carry currents of equal magnitude I.They repel each other with a force per unit length of 5.8 nN/m.The current I is approximately

Ampère's law is valid

Two long,straight parallel wires 7.6 cm apart carry currents of equal magnitude I.They attract each other with a force per unit length of 4.3 nN/m.The current I is approximately

If the magnetic field at the center of a square current loop of side L = 40 cm is 2.4 ×\times× 10-6 T calculate the current flowing around the loop.

Use the following to answer the next question: -Two long parallel wires are a distance d apart (d = 6 cm) and carry equal and opposite currents of 5 A.Point P is distance d from each of the wires.Calculate the magnitude of the magnetic field strength at point P.

A long solenoid of radius 3 cm has 1100 turns per m.If the solenoid wire carries a current of 1.5 A,then calculate the magnetic field inside the solenoid.

Two straight rods 55 cm long and 1.4 mm apart in a current balance carry current I,each in opposite directions.If a mass of 1.3 g when placed on the upper rod balances the magnetic force of repulsion between the rods,then calculate the current I.

A coil of radius R = 5 cm lies inside a long solenoid which carries a current I1 = 10 A and has 800 turns/m.The coil carries a current I2 = 4 A and the normal of the plane of the coil is at an angle θ\thetaθ = 30 ∘\circ∘ with the axis of the solenoid.The magnitude of the torque on the coil is

Two parallel wires carry currents I1 and I2 = 2I1 in the same direction.Which of the following expressions shows the relationship of the forces on the wires?

Gauss's law for magnetism summarizes the fact(s) that

Calculate the magnetic field at the center of a circular current loop of area 0.200 m² divided by the magnetic field at the center of a square current loop of the same inside area for the same current I value.

If the magnetic field at the center of a square current loop of side L = 40 cm is 2.4 $\times$ 10^-⁶ T calculate the current flowing around the loop.

A coil of radius R = 5 cm lies inside a long solenoid which carries a current I₁ = 10 A and has 800 turns/m.The coil carries a current I₂ = 4 A and the normal of the plane of the coil is at an angle $\theta$ = 30^$\circ$ with the axis of the solenoid.The magnitude of the torque on the coil is

Two parallel wires carry currents I₁ and I₂ = 2I₁ in the same direction.Which of the following expressions shows the relationship of the forces on the wires?