Question 1

A "coulomb" is: A) one ampere per second B) the quantity of charge which will exert a force of 1 N on a similar charge at a distance of 1 m C) the amount of current in each of two long parallel wires separated by 1 m, which produces a force of 2 * 10^-⁷ N per meter D) the amount of charge which flows past a point in one second when the current is 1 A E) an abbreviation for a certain combination of kilogram, meter and second

Accepted Answer

A coulomb is defined as the amount of electric charge that flows past a point in a circuit in one second when the current is one ampere.

Question 2

In the figure, the current element  the point P, and the three vectors (1, 2, 3) are all in the plane of the page. The direction of  due to this current element, at the point P is: 
A) in the direction marked "1"
B) in the direction marked "2"
C) in the direction marked "3"
D) out of the page
E) into the page

Accepted Answer

E

Question 3

Two long straight wires pierce the plane of the paper at vertices of an equilateral triangle as shown below. They each carry 2 A, out of the paper. The magnetic field at the third vertex (P) has magnitude (in T): A) 1.0 *10^-⁵ B) 1.7* 10^-⁵ C) 2.0 * 10^-⁵ D) 5.0 * 10^-⁶ E) 8.7 *10^-⁶

Accepted Answer

We can use the equation for the magnetic field due to a current-carrying wire, which is given by:
B = (μ0/4π) * (I/d)
where μ0 is the permeability of free space, I is the current, and d is the distance from the wire to the point where the magnetic field is being measured.

Let's first consider the wire that is closest to point P. The distance from this wire to point P is equal to one side of the equilateral triangle, which we can call a. Therefore, the magnetic field due to this wire at point P is given by:
B1 = (μ0/4π) * (2/ a)

Now consider the other wire. To find the magnetic field due to this wire at point P, we can use the fact that the angle between the line connecting this wire to point P and the plane of the equilateral triangle is equal to 60 degrees. Therefore, the distance from this wire to point P is equal to (a/2) / sin(60), which simplifies to a/√3. Thus, the magnetic field due to this wire at point P is given by:
B2 = (μ0/4π) * (2/ (a/√3)) = (μ0/4π) * (6/ a)

The net magnetic field at point P is the vector sum of the magnetic fields due to each wire. Since the wires are carrying current in the same direction, their magnetic fields will add together. Therefore, the total magnetic field at point P is given by:
B = B1 + B2 = (μ0/4π) * (2/a) + (μ0/4π) * (6/a) = (μ0/4π) * (8/a)

Now we just need to plug in the given values and solve:
B = (μ0/4π) * (8/0.1) = 1.7 T

Therefore, the best choice is (B).

Question 4

The diagram shows three equally spaced wires that are perpendicular to the page. The currents are all equal, two being out of the page and one being into the page. Rank the wires according to the magnitudes of the magnetic forces on them, from least to greatest. 
A) 1, 2, 3
B) 2, 1 and 3 tie
C) 2 and 3 tie, then 1
D) 1 and 3 tie, then 2
E) 3, 2, 1

Accepted Answer

The answer of The diagram shows three equally spaced wires...

Question 5

Suitable units for $\mu$₀ are: A) tesla B) newton/ampere² C) weber/meter D) kilogram.ampere/meter E) tesla.meter/ampere

Accepted Answer

The given expression involves units of capacitance (microfarads), magnetic field strength (teslas), and electric current per unit length (amperes per meter). The only option that includes all of these units is E, which expresses the units as teslas multiplied by meters per ampere.

Question 6

A wire carrying a large current i from east to west is placed over an ordinary magnetic compass. The end of the compass needle marked "N" will point:
A) north
B) south
C) east
D) west
E) the compass will act as an electric motor, hence the needle will keep rotating

Accepted Answer

According to the right-hand rule of electromagnetism, when a current flows through a wire, a magnetic field is created around the wire in a direction perpendicular to the current flow. This magnetic field will interact with the Earth's magnetic field, causing the compass needle to deflect. The direction of deflection is perpendicular to both the direction of the current flow (east to west) and the Earth's magnetic field (north to south). Therefore, the end of the compass needle marked "N" will point south (as it is opposite to the direction of the deflection).

Question 7

Two long parallel straight wires carry equal currents in opposite directions. At a point midway between the wires, the magnetic field they produce is:
A) zero
B) non-zero and along a line connecting the wires
C) non-zero and parallel to the wires
D) non-zero and perpendicular to the plane of the two wires
E) none of the above

Accepted Answer

By applying the right-hand rule for the magnetic field around a wire, we can see that each wire produces a magnetic field that is perpendicular to the plane of the wire and circles around it. At the midpoint between the wires, these two magnetic fields are in opposite directions and add up to create a net magnetic field that is perpendicular to the plane of the wires. Therefore, choice D is correct. The other choices are incorrect because there is a non-zero magnetic field present at the midpoint between the wires.

Question 8

Electrons are going around a circle in a counterclockwise direction as shown. At the center of the circle they produce a magnetic field that is: 
A) into the page
B) out of the page
C) to the left
D) to the right
E) zero

Accepted Answer

The answer of Electrons are going around a circle in...

Question 9

Two long straight current-carrying parallel wires cross the x axis and carry currents I and 3I in the same direction, as shown. At what value of x is the net magnetic field zero? 
A) 0
B) 1
C) 3
D) 5
E) 7

Accepted Answer

The magnetic field produced by a wire carrying current I at a distance r from the wire is given by:
$B = \frac{\mu_0 I}{2\pi r}$, where $\mu_0$ is the permeability of free space.
For wire carrying current 3I, the magnetic field produced at distance r is:
$B' = \frac{\mu_0 (3I)}{2\pi r} = 3\frac{\mu_0 I}{2\pi r} = 3B$
At any point on the x axis, the magnetic field due to the wire carrying current I is in the opposite direction to the magnetic field due to the wire carrying current 3I, so the net magnetic field is given by:
$B_{net} = B - B' = B - 3B = -2B$
For the net magnetic field to be zero, we need $B=0$, which occurs when $x=3$. Therefore, the answer is C.

Question 10

Which graph correctly gives the magnitude of the magnetic field outside an infinitely long straight current-carrying wire as a function of the distance r from the wire? 
A) I
B) II
C) III
D) IV
E) V

Accepted Answer

The answer of Which graph correctly gives the magnitude of...

Question 11

Two long straight wires are parallel and carry current in opposite directions. The currents are 8.0 A and 12 A and the wires are separated by 0.40 cm. The magnetic field in tesla at a point midway between the wires is: A) 0 B) 4.0 * 10^-⁴ C) 8.0 * 10^-⁴ D) 12 * 10^-⁴ E) 20 * 10^-⁴

Accepted Answer

The answer of Two long straight wires are parallel and...

Question 12

In an overhead straight wire, the current is north. The magnetic field due to this current, at our point of observation, is:
A) east
B) up
C) north
D) down
E) west

Accepted Answer

Using the right-hand rule, if the current is flowing north, the magnetic field at a point to the right of the wire (assuming we are facing the same direction as the current) will be directed towards the west.

Question 13

The magnetic field a distance 2 cm from a long straight current-carrying wire is 2 *10^-⁵ T. The current in the wire is: A) 0.16 A B) 1.0 A C) 2.0 A D) 4.0 A E) 25 A

Accepted Answer

From the formula for magnetic field around a long straight current-carrying wire, we have:

B = μ0I/2πr, where μ0 = 4π×10^-7 T·m/A is the permeability of free space, I is the current in amperes, and r is the distance from the wire in meters.

Rearranging the formula, we have:

I = 2πrB/μ0

Substituting r = 2 cm = 0.02 m and B = 2×10^-5 T, we get:

I = 2π×0.02×2×10^-5/4π×10^-7 = 2 A

Therefore, the current in the wire is 2.0 A, which is answer choice C.

Question 14

The magnitude of the magnetic field at point P, at the center of the semicircle shown, is given by: A) 2$\mu$₀_i/R² B) $\mu$₀i/2 $\pi$R C) $\mu$₀i/4 $\pi$R D) $\mu$₀i/2R E) $\mu$₀i/4R

Accepted Answer

The answer of The magnitude of the magnetic field at...

Question 15

Lines of the magnetic field produced by a long straight wire carrying a current are:
A) in the direction of the current
B) opposite to the direction of the current
C) leave the wire radially
D) are circles concentric with the wire
E) are lines similar to those produced by a bar magnet

Accepted Answer

The answer of Lines of the magnetic field produced by...

Question 16

Two parallel long wires carry the same current and repel each other with a force F per unit length. If both these currents are doubled and the wire separation tripled, the force per unit length becomes:
A) 2F/9
B) 4F/9
C) 2F/3
D) 4F/3
E) 6F

Accepted Answer

The answer of Two parallel long wires carry the same...

Question 17

Two long straight wires are parallel and carry current in the same direction. The currents are 8.0 and 12 A and the wires are separated by 0.40 cm. The magnetic field in tesla at a point midway between the wires is: A) 0 B) 4.0 *10^-⁴ C) 8.0 * 10^-⁴ D) 12 *10^-⁴ E) 20 * 10^-⁴

Accepted Answer

Using the formula for the magnetic field due to a straight wire, we have:
B1 = μ0I1/(2πr)
B2 = μ0I2/(2πr)
The total magnetic field at the midpoint between the wires is the sum of the magnetic fields due to each wire:
B = B1 + B2 = μ0I1/(2πr) + μ0I2/(2πr)
B = (μ0/2πr)(I1 + I2)
B = (4π×10^-7 T·m/A)(20 A)/(0.004 m)
B ≈ 4.0×10^-4 T ≈ 0.40 mT
Therefore, the best choice is B) 4.0×10^-4 T.

Question 18

The magnetic field outside a long straight current-carrying wire depends on the distance R from the wire axis according to: A) R B) 1/R C) 1/R² D) 1/R³ E) 1/R^3/2

Accepted Answer

The magnetic field outside a long straight current-carrying wire varies inversely with the distance from the wire axis, so the correct answer is 1/R.

Question 19

The diagrams show three circuits consisting of concentric circular arcs (either half or quarter circles of radii r, 2r, and 3r) and radial lengths. The circuits carry the same current. Rank them according to the magnitudes of the magnetic fields they produce at C, least to greatest. 
A) 1, 2, 3
B) 3, 2, 1
C) 1, 3, 2
D) 2, 3, 1
E) 2, 1, 3

Accepted Answer

The answer of The diagrams show three circuits consisting of...

Question 20

Two parallel wires carrying equal currents of 10 A attract each other with a force of 1 mN. If both currents are doubled, the force of attraction will be:
A) 1 mN
B) 4 mN
C) 0.5 mN
D) 0.25 mN
E) 2 mN

Accepted Answer

The force of attraction between the two wires is given by the formula:

F = (μ₀*I₁*I₂*L)/(2π*r)

where F is the force of attraction, μ₀ is the permeability of free space, I₁ and I₂ are the currents, L is the length of the wires, and r is the distance between the wires.

Since the currents in both wires are doubled to 20 A, the force of attraction between the wires will be:

F' = (μ₀*20 A*20 A*L)/(2π*r)

= 4*(μ₀*10 A*10 A*L)/(2π*r)

= 4*F

Therefore, the force of attraction between the wires will be 4 times the original force, which is 1 mN. So the answer is B.

Quiz 29: Magnetic Fields Due to Currents

A "coulomb" is:

In the figure, the current element the point P, and the three vectors (1, 2, 3) are all in the plane of the page. The direction of due to this current element, at the point P is:

Two long straight wires pierce the plane of the paper at vertices of an equilateral triangle as shown below. They each carry 2 A, out of the paper. The magnetic field at the third vertex (P) has magnitude (in T) :

The diagram shows three equally spaced wires that are perpendicular to the page. The currents are all equal, two being out of the page and one being into the page. Rank the wires according to the magnitudes of the magnetic forces on them, from least to greatest.

Suitable units for $\mu$ ₀ are:

A wire carrying a large current i from east to west is placed over an ordinary magnetic compass. The end of the compass needle marked "N" will point:

Two long parallel straight wires carry equal currents in opposite directions. At a point midway between the wires, the magnetic field they produce is:

Electrons are going around a circle in a counterclockwise direction as shown. At the center of the circle they produce a magnetic field that is:

Two long straight current-carrying parallel wires cross the x axis and carry currents I and 3I in the same direction, as shown. At what value of x is the net magnetic field zero?

Which graph correctly gives the magnitude of the magnetic field outside an infinitely long straight current-carrying wire as a function of the distance r from the wire?

Two long straight wires are parallel and carry current in opposite directions. The currents are 8.0 A and 12 A and the wires are separated by 0.40 cm. The magnetic field in tesla at a point midway between the wires is:

In an overhead straight wire, the current is north. The magnetic field due to this current, at our point of observation, is:

The magnetic field a distance 2 cm from a long straight current-carrying wire is 2 *10^-⁵ T. The current in the wire is:

Lines of the magnetic field produced by a long straight wire carrying a current are:

Two parallel long wires carry the same current and repel each other with a force F per unit length. If both these currents are doubled and the wire separation tripled, the force per unit length becomes:

Two long straight wires are parallel and carry current in the same direction. The currents are 8.0 and 12 A and the wires are separated by 0.40 cm. The magnetic field in tesla at a point midway between the wires is:

The magnetic field outside a long straight current-carrying wire depends on the distance R from the wire axis according to:

The diagrams show three circuits consisting of concentric circular arcs (either half or quarter circles of radii r, 2r, and 3r) and radial lengths. The circuits carry the same current. Rank them according to the magnitudes of the magnetic fields they produce at C, least to greatest.

Two parallel wires carrying equal currents of 10 A attract each other with a force of 1 mN. If both currents are doubled, the force of attraction will be: