Question 1

The magnetic field at a distance of 2 cm from a current carrying wire is 4 μT. What is the magnetic field at a distance of 4 cm from the wire?

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Question 2

A wire segment 1.2 m long carries a current I = 3.5 A, and is oriented as shown in the figure. The +x-axis points directly into the page. A uniform magnetic field of magnitude 0.50 T pointing toward the -x direction is present as shown. What is the magnetic force vector on the wire segment?

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Question 3

A wire carries a 4.0-A current along the +x-axis through a magnetic field  = (5.0 
+ 7.0 
) T. If the wire experiences a force of 30 N 
As a result, how long is the wire?

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Question 4

An L-shaped metal machine part is made of two equal-length segments that are perpendicular to each other and carry a 4.50-A current as shown in the figure. This part has a total mass of 3.80 kg and a total length of 3.00 m, and it is in an external 1.20-T magnetic field that is oriented perpendicular to the plane of the part, as shown. What is the magnitude of the NET magnetic force that the field exerts on the part?

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Question 5

A very long thin wire produces a magnetic field of 0.0050 × 10^-4 T at a distance of 3.0 mm. from the central axis of the wire. What is the magnitude of the current in the wire? (μ₀ = 4π × 10^-7 T ∙ m/A)

Accepted Answer

The magnetic field of a long, thin wire is given by:$$B = \frac{\mu_0 I}{2\pi r}$$where:* B is the magnetic field* μ0 is the permeability of free space (4π × 10^-7 T ∙ m/A)* I is the current in the wire* r is the distance from the central axis of the wireWe can rearrange this equation to solve for the current:$$I = \frac{2\pi rB}{\mu_0}$$Plugging in the given values, we get:$$I = \frac{2\pi (3.0 	imes 10^{-3} 	ext{ m})(0.0050 	imes 10^{-4} 	ext{ T})}{4\pi 	imes 10^{-7} 	ext{ T} \cdot 	ext{m/A}} = 7.5 	imes 10^{-3} 	ext{ A} = 7.5 	ext{ mA}$$Therefore, the magnitude of the current in the wire is 7.5 mA.

Question 6

A straight 15.0-g wire that is 2.00 m long carries a current of 8.00 A. This wire is aligned horizontally along the west-east direction with the current going from west to east. You want to support the wire against gravity using the weakest possible uniform external magnetic field.
(a) Which way should the magnetic field point?
(b) What is the magnitude of the weakest possible magnetic field you could use?

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Question 7

A rectangular loop of wire carrying a 4.0-A current is placed in a magnetic field of 0.60 T. The magnitude of the torque acting on this wire when the plane of the loop makes a 30° angle with the field is measured to be 1.1 N ∙ m. What is the area of this loop?

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Question 8

A rigid circular loop has a radius of 0.20 m and is in the xy-plane. A clockwise current I is carried by the loop, as shown. The magnitude of the magnetic moment of the loop is 0.75 A · m². A uniform external magnetic field, B = 0.20 T in the positive x-direction, is present. (a) What is the current in the loop? (b) Find the magnitude of the magnetic torque exerted on the loop. (c) If the loop is released from rest, in what direction will points a and c initially move?

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Question 9

A 15-turn rectangular loop of wire of width 10 cm and length 20 cm has a current of 2.5 A flowing through it. Two sides of the loop are oriented parallel to a uniform magnetic field of strength 0.037 T, and the other two sides are perpendicular to the magnetic field.
(a) What is the magnitude of the magnetic moment of the loop?
(b) What torque does the magnetic field exert on the loop?

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Question 10

A circular loop of diameter 10 cm, carrying a current of 0.20 A, is placed inside a magnetic field  = 0.30 T 
. The normal to the loop is parallel to a unit vector 
= -0.60 
- 0.80 
. Calculate the magnitude of the torque on the loop due to the magnetic field.

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Question 11

A wire along the z-axis carries a current of 6.8 A in the +z direction. Find the magnitude and direction of the force exerted on a 6.1-cm long length of the wire by a uniform magnetic field with magnitude 0.36 T in the -x direction.

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Question 12

A thin copper rod that is 1.0 m long and has a mass of 0.050 kg is in a magnetic field of 0.10 T. What minimum current in the rod is needed in order for the magnetic force to cancel the weight of the rod?

Accepted Answer

The weight of the rod is given by:
$$F_g=mg=0.050~	ext{kg}\cdot 9.81~	ext{m/s}^2=0.4905~	ext{N}$$
The magnetic force on a current-carrying conductor in a magnetic field is given by:
$$F_B=ILB\sin	heta$$
where $I$ is the current in the rod, $L$ is the length of the rod, $B$ is the magnetic field, and $	heta$ is the angle between the magnetic field and the direction of the current. In this case, the magnetic force must cancel the weight of the rod, so we have:
$$ILB\sin	heta=F_g$$
Solving for $I$:
$$I=\frac{F_g}{LB\sin	heta}$$
Since $\sin	heta=1$ for a current that is perpendicular to the magnetic field, we have:
$$I=\frac{F_g}{LB}=\frac{0.4905~	ext{N}}{1.0~	ext{m}\cdot 0.10~	ext{T}}=4.905~	ext{A}$$
Therefore, the minimum current needed to cancel the weight of the rod is 4.9 A, which is choice C.

Question 13

A circular coil of wire of 200 turns and diameter 2.0 cm carries a current of 4.0 A. It is placed in a magnetic field of 0.70 T with the plane of the coil making an angle of 30° with the magnetic field. What is the magnetic torque on the coil?

Accepted Answer

The magnetic torque on a current loop is given by the equation:$$	au = NIAB\sin	heta$$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 magnetic field and the normal to the coilIn this problem, we have:* N = 200 turns* I = 4.0 A* A = πr² = π(0.01 m)² = 3.14 x 10^-4 m²* B = 0.70 T* θ = 30°Plugging these values into the equation, we get:$$	au = (200)(4.0 A)(3.14 x 10^-4 m²)(0.70 T)\sin 30° = 0.15 N ∙ m$$Therefore, the correct answer is A.

Question 14

A point charge Q moves on the x-axis in the positive direction with a speed of 370 m/s. A point P is on the y-axis at y = +80 mm. The magnetic field produced at point P, as the charge moves through the origin, is equal to When the charge is at What is the magnitude of the magnetic field at point P? (μ₀ = 4π × 10^-7 T ∙ m/A)

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Question 15

A wire segment 1.2 m long carries a current I = 3.5 A and is oriented as shown in the figure. A uniform magnetic field of magnitude 0.50 T pointing toward the -x direction is present as shown. The +z-axis points directly into the page. What is the magnetic force vector on the wire segment?

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Question 16

A wire in the shape of an "M" lies in the plane of the paper. It carries a current of 2.0 A, flowing from points A to E, as shown in the figure. It is placed in a uniform magnetic field of 0.75 T in the same plane, directed as shown on the right side of the figure. The figure indicates the dimensions of the wire. What are the magnitude and direction of the force acting on  
(a) section AB of this wire?
(b) section BC of this wire?
(c) section CD of this wire?
(d) section DE of this wire?
(e) the entire wire?

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Question 17

A point charge Q moves on the x-axis in the positive direction with a speed of 280 m/s. A point P is on the y-axis at y = +70 mm. The magnetic field produced at the point P, as the charge moves through the origin, is equal to What is the charge Q? (μ₀ = 4π × 10^-7 T ∙ m/A)

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Question 18

A rigid rectangular loop, which measures 0.30 m by 0.40 m, carries a current of 5.5 A, as shown in the figure. A uniform external magnetic field of magnitude 2.9 T in the negative x direction is present. Segment CD is in the xz-plane and forms a 35° angle with the z-axis, as shown. Find the magnitude of the external torque needed to keep the loop in static equilibrium.

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Question 19

A straight wire that is 0.60 m long is carrying a current of 2.0 A. It is placed in a uniform magnetic field of strength 0.30 T. If the wire experiences a force of 0.18 N, what angle does the wire make with respect to the magnetic field?

Accepted Answer

The force on a current-carrying wire in a magnetic field is given by F = ILBsinθ, where I is the current, L is the length of the wire in the magnetic field, B is the strength of the magnetic field, and θ is the angle between the wire and the magnetic field. Solving for θ, we get sinθ = F/ILB = (0.18 N)/(2.0 A)(0.60 m)(0.30 T) = 0.50, which corresponds to θ = 30°. Therefore, the answer is B.

Question 20

At what distance from the central axis of a long straight thin wire carrying a current of 5.0 A is the magnitude of the magnetic field due to the wire equal to the strength of the Earth's magnetic field of about 5.0 × 10^-5 T? (μ₀ = 4π × 10^-7 T ∙ m/A)

Accepted Answer

We can use the formula for the magnetic field due to a long straight wire, given by:

B = μ₀I/2πr

Where B is the magnetic field, μ₀ is the vacuum permeability (given as μ₀ = 4π × 10^-7 T · m/A), I is the current, and r is the distance from the wire.

We want to find the distance at which the magnetic field B is equal to the strength of the Earth's magnetic field, given as B = 5.0 × 10^-5 T.

Setting these two expressions equal, we get:

5.0 × 10^-5 T = μ₀I/2πr

Solving for r, we get:

r = μ₀I/2πB

Plugging in the given values, we get:

r = (4π × 10^-7 T · m/A)(5.0 A)/(2π)(5.0 × 10^-5 T) ≈ 2.0 cm

Therefore, the answer is B) 2.0 cm.

Quiz 32: The Magnetic Field

The magnetic field at a distance of 2 cm from a current carrying wire is 4 μT. What is the magnetic field at a distance of 4 cm from the wire?

A wire segment 1.2 m long carries a current I = 3.5 A, and is oriented as shown in the figure. The +x-axis points directly into the page. A uniform magnetic field of magnitude 0.50 T pointing toward the -x direction is present as shown. What is the magnetic force vector on the wire segment?

A wire carries a 4.0-A current along the +x-axis through a magnetic field = (5.0 + 7.0 ) T. If the wire experiences a force of 30 N As a result, how long is the wire?

A very long thin wire produces a magnetic field of 0.0050 × 10-4 T at a distance of 3.0 mm. from the central axis of the wire. What is the magnitude of the current in the wire? (μ0 = 4π × 10-7 T ∙ m/A)