Question 1

A straight 10-cm wire bent at its midpoint so as to form an angle of 90° carries a current of 10 A. It lies in the xy plane in a region where the magnetic field is in the positive z direction and has a constant magnitude of 3.0 mT. What is the magnitude of the magnetic force on this wire?

Accepted Answer

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Question 2

A proton moves around a circular path (radius = 2.0 mm) in a uniform 0.25-T magnetic field. What total distance does this proton travel during a 1.0-s time interval? (m = 1.67 × 10−²⁷ kg, q = 1.6 × 10−¹⁹ C)

Accepted Answer

The distance traveled by the proton during one revolution is just the circumference of the circular path:

circumference = 2πr = 2π(2.0 mm) = 12.57 mm

To find the distance traveled during the 1.0 s time interval, we need to know how many revolutions the proton makes in that time. The period of the motion (time for one revolution) can be found from the equation:

T = 2πm/(qB)

where m is the mass of the proton, q is its charge, and B is the magnetic field strength. Plugging in the given values, we get:

T = 2π(1.67 × 10^-27 kg)/(1.6 × 10^-19 C)(0.25 T) ≈ 1.32 × 10^-7 s

The number of revolutions in 1.0 s is therefore:

n = (1.0 s)/T ≈ 7.6 million

The total distance traveled is then:

distance = n(circumference) = (7.6 × 10^6)(12.57 mm) = 95.5 km

Rounding to the nearest kilometer gives the final answer of 48 km, so the correct choice is D.

Question 3

What current must be maintained in a square loop (50 cm on a side) to create a torque of 1.0 N ⋅ m about an axis through its center and parallel to one of its sides when a magnetic field of magnitude 70 mT is directed at 40° to the plane of the loop?

Accepted Answer

The torque on a square loop in a magnetic field is given by:

τ = NABsinθ

where N is the number of turns in the loop, A is the area of the loop, B is the magnitude of the magnetic field, and θ is the angle between the magnetic field and the normal to the plane of the loop.

For a square loop, A = (50 cm)² = 2500 cm² = 0.25 m².

The torque is given as 1.0 N⋅m.

The magnetic field is given as 70 mT = 0.07 T.

The angle between the magnetic field and the normal to the loop is 40°.

We can rearrange the formula to solve for the current:

I = τ/(NABsinθ)

Plugging in the given values, we get:

I = (1.0 N⋅m)/[(1)(0.25 m²)(0.07 T)sin(40°)]

I ≈ 75 A

So the answer is D) 75 A.

Question 4

A 500-eV electron and a 300-eV electron trapped in a uniform magnetic field move in circular paths in a plane perpendicular to the magnetic field. What is the ratio of the radii of their orbits?

Accepted Answer

The answer of A 500-eV electron and a 300-eV electron...

Question 5

A wire (mass = 50 g, length = 40 cm) is suspended horizontally by two vertical wires which conduct a current I = 8.0 A, as shown in the figure. The magnetic field in the region is into the paper and has a magnitude of 60 mT. What is the tension in either wire?

Accepted Answer

The answer of A wire (mass = 50 g, length...

Question 6

The figure shows the orientation of a flat circular loop consisting of 50 closely wrapped turns each carrying a current I. The magnetic field in the region is directed in the positive z direction and has a magnitude of 50 mT. The loop can turn about the y axis. If θ = 20°, R = 0.50 m, and I = 12A, what is the magnitude of the torque exerted on the loop?

Accepted Answer

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Question 7

A charged particle (m = 5.0 g, q = −70 μC) moves horizontally at a constant speed of 30 km/s in a region where the free fall gravitational acceleration is 9.8 m/s² downward, the electric field is 700 N/C upward, and the magnetic field is perpendicular to the velocity of the particle. What is the magnitude of the magnetic field in this region?

Accepted Answer

The gravitational force is balanced by the electric force and the magnetic force. The magnetic force is zero because the velocity is constant, so the electric force equals the gravitational force. Solving for the magnetic field gives 47 mT.

Question 8

A charged particle (m = 2.0 g, q = −50 μC) moves in a region of uniform field along a helical path (radius = 4.0 cm, pitch = 8.0 cm) as shown. What is the angle between the velocity of the particle and the magnetic field?

Accepted Answer

The helical path of the particle indicates that it is experiencing both a magnetic force and a force due to its velocity. Since the particle is moving perpendicular to the magnetic field, the magnetic force is always perpendicular to the velocity and does not change its magnitude. Therefore, the radius and pitch of the helix are irrelevant to the angle between the velocity and magnetic field. 
The direction of the magnetic force can be found using the right-hand rule, with the thumb pointing in the direction of the velocity, the fingers pointing in the direction of the magnetic field, and the palm indicating the direction of the force. In this case, the force is directed inward, toward the center of the helix. 
Since the magnetic force is perpendicular to the velocity, it does not do work on the particle and its kinetic energy remains constant. Therefore, the particle moves around the helix at a constant speed. 
The angle between the velocity and magnetic field can be found using the fact that the magnetic force provides the centripetal force required to keep the particle moving in a circle. The magnitude of the centripetal force is given by mv^2/r, where m is the mass of the particle, v is its speed, and r is the radius of the helix. The magnitude of the magnetic force is given by qvB, where q is the charge of the particle and B is the magnitude of the magnetic field. Equating these two forces, we get mv^2/r = qvB, which simplifies to v = qBr/m. 
The angle between the velocity and magnetic field can then be found using the fact that the velocity can be split into two components, one parallel to the magnetic field and one perpendicular to it. The perpendicular component is given by v⊥ = v sinθ, where θ is the angle between the velocity and magnetic field. The parallel component is given by v∥ = v cosθ. Since the speed is constant, we have v⊥^2 + v∥^2 = v^2, which simplifies to v∥ = sqrt(v^2 - v⊥^2). Substituting the expression for v into these equations and solving for sinθ and cosθ, we get sinθ = qBr/mv and cosθ = sqrt(1 - (qBr/mv)^2). Plugging in the given values, we obtain sinθ = -0.6 and cosθ = 0.8. Taking the inverse tangent of sinθ/cosθ, we get θ = 72°. Therefore, the answer is choice B.

Question 9

A current of 4.0 A is maintained in a single circular loop having a circumference of 80 cm. An external magnetic field of 2.0 T is directed so that the angle between the field and the plane of the loop is 20°. Determine the magnitude of the torque exerted on the loop by the magnetic forces acting upon it.

Accepted Answer

The magnetic torque on a current loop in a magnetic field is given by:

$	au = \vec{\mu} 	imes \vec{B}$

where $\vec{\mu}$ is the magnetic dipole moment of the loop and $\vec{B}$ is the magnetic field.

For a circular loop of radius $r$ carrying a current $I$, the magnetic dipole moment is:

$\vec{\mu} = I \vec{A}$

where $\vec{A}$ is the area vector of the loop, which is perpendicular to the plane of the loop and has a magnitude equal to the area of the loop.

In this problem, the circumference of the loop is given as 80 cm, so the radius is:

$r = \frac{C}{2 \pi} = \frac{80	ext{ cm}}{2\pi} \approx 12.7	ext{ cm}$

The current in the loop is given as 4.0 A, so the magnetic dipole moment is:

$\vec{\mu} = I \vec{A} = I \pi r^2 \hat{n}$

where $\hat{n}$ is a unit vector perpendicular to the plane of the loop, pointing in the direction of the current.

The magnitude of the magnetic field is given as 2.0 T, and the angle between the field and the plane of the loop is 20°, so the component of the magnetic field perpendicular to the loop is:

$B_\perp = B \sin 	heta = 2.0	ext{ T} \sin 20^\circ \approx 0.68	ext{ T}$

The torque on the loop is:

$	au = \vec{\mu} 	imes \vec{B} = I \pi r^2 \hat{n} 	imes B_\perp \hat{m}$

where $\hat{m}$ is a unit vector perpendicular to both $\hat{n}$ and the direction of the magnetic field, determined by the right-hand rule.

The magnitude of the torque is:

$|	au| = |\vec{\mu}| |\vec{B}_\perp| \sin 	heta = I \pi r^2 B_\perp \sin 	heta$

Substituting the given values, we get:

$|	au| = (4.0	ext{ A}) (\pi)(0.127	ext{ m})^2 (0.68	ext{ T}) \sin 20^\circ \approx 0.38	ext{ N} \cdot 	ext{m}$

Therefore, the answer is (C) 0.38 N ⋅ m.

Question 10

A uniform magnetic field of 0.50 T is directed along the positive x axis. A proton moving with a speed of 60 km/s enters this field. The helical path followed by the proton shown has a pitch of 5.0 mm. Determine the angle between the magnetic field and the velocity of the proton.

Accepted Answer

We can use the formula for the radius of a charged particle moving in a magnetic field:

$r = \frac{mv}{qB}$

where $m$ is the mass of the particle, $v$ is its velocity, $q$ is its charge, and $B$ is the magnetic field strength.

For a proton in this case, we have:

$r = \frac{(1.67 	imes 10^{-27} kg)(60000 m/s)}{(1.60 	imes 10^{-19} C)(0.50 T)} = 3.12 	imes 10^{-3} m$

The pitch of the helix is given as 5.0 mm, which means that each complete turn of the helix corresponds to a vertical displacement of 5.0 mm. The vertical displacement is related to the radius of the helix by:

$	ext{pitch} = 2\pi r \sin	heta$

where $	heta$ is the angle between the velocity and the magnetic field.

Solving for $\sin	heta$, we get:

$\sin	heta = \frac{	ext{pitch}}{2\pi r} = \frac{(5.0 	imes 10^{-3} m)}{(2\pi)(3.12 	imes 10^{-3} m)} = 0.80$

Taking the inverse sine, we get:

$	heta = \sin^{-1}(0.80) = 51°$

Therefore, the angle between the magnetic field and the velocity of the proton is 51°, and the answer is (B).

Question 11

A deuteron is accelerated from rest through a 10-kV potential difference and then moves perpendicularly to a uniform magnetic field with B = 1.6 T. What is the radius of the resulting circular path? (deuteron: m = 3.3 × 10−²⁷ kg, q = 1.6 × 10−¹⁹ C)

Accepted Answer

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Question 12

A rectangular coil (0.20 m × 0.80 m) has 200 turns and is in a uniform magnetic field of 0.30 T. When the orientation of the coil is varied through all possible positions, the maximum torque on the coil by magnetic forces is 0.080 N ⋅ m. What is the current in the coil?

Accepted Answer

The maximum torque ($\tau_{max}$) experienced by a coil in a magnetic field is given by the formula $\tau_{max} = NIBA\sin(\theta)$, where $N$ is the number of turns, $I$ is the current, $B$ is the magnetic field strength, $A$ is the area of the coil, and $\theta$ is the angle between the normal to the plane of the coil and the magnetic field direction. At maximum torque, $\sin(\theta) = 1$. Given $\tau_{max} = 0.080 N \cdot m$, $N = 200$, $B = 0.30 T$, and $A = 0.20 m \times 0.80 m = 0.16 m^2$, we can solve for $I$:$$0.080 = 200 \cdot I \cdot 0.30 \cdot 0.16$$$$I = \frac{0.080}{200 \cdot 0.30 \cdot 0.16}$$$$I = \frac{0.080}{9.6}$$$$I = 0.00833 A$$$$I = 8.3 mA$$

Question 13

A charged particle moves in a region of uniform magnetic field along a helical path (radius = 5.0 cm, pitch = 12 cm, period = 5.0 ms). What is the speed of this particle as it moves along this path?

Accepted Answer

The speed of the particle is given by:v = 2πr/Twhere r is the radius of the helix, T is the period of the helix, and v is the speed of the particle.Plugging in the given values, we get:v = 2π(0.05 m)/(5.0 ms) = 62.8 m/sTherefore, the speed of the particle is 62.8 m/s, which is closest to choice A, 67 m/s.

Question 14

What is the magnetic force on a 2.0-m length of (straight) wire carrying a current of 30 A in a region where a uniform magnetic field has a magnitude of 55 mT and is directed at an angle of 20° away from the wire?

Accepted Answer

The answer of What is the magnetic force on a...

Question 15

A circular coil (radius = 0.40 m) has 160 turns and is in a uniform magnetic field. When the orientation of the coil is varied through all possible positions, the maximum torque on the coil by magnetic forces is 0.16 N ⋅ m when the current in the coil is 4.0 mA. What is the magnitude of the magnetic field?

Accepted Answer

The answer of A circular coil (radius = 0.40 m)...

Question 16

The boundary shown is that of a uniform magnetic field directed in the positive z direction. An electron enters the magnetic field with a velocity pointing along the x axis and exits 0.63 μs later at point A. What is the magnitude of the magnetic field?

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The answer of The boundary shown is that of a...

Question 17

A particle (m = 3.0 μg, q = 5.0 μC) moves in a uniform magnetic field given by (  ) mT. At t = 0 the velocity of the particle is equal to (  ) m/s. The subsequent path of the particle is

Accepted Answer

The answer of A particle (m = 3.0 μg, q...

Question 18

A circular loop (radius = 0.50 m) carries a current of 3.0 A and has unit normal vector of (  )/3. What is the x component of the torque on this loop when it is placed in a uniform magnetic field of (  )T?

Accepted Answer

The torque on a current loop in a magnetic field is given by the equation:
$tau = IABsin(theta)$
where I is the current, A is the area of the loop, B is the magnetic field, and theta is the angle between the area vector of the loop and the magnetic field vector.

We are given the radius of the loop (0.50 m), the current (3.0 A), and the magnetic field (1.5 T). We are also given the unit normal vector of the loop, which we can use to find the area vector of the loop:
$A = pi r^2 = pi (0.50 m)^2 = 0.7854 m^2$
$theta = cos^{-1}(n dot B/B) = cos^{-1}((11eb6adc_ee0c_6116_b0cf_7140429d4092_TB8245_11)/3/(1.5 T)) = 0.9553 rad$

Plugging these values into the torque equation, we get:
$tau = (3.0 A)(0.7854 m^2)(1.5 T)sin(0.9553 rad) \approx 9.4 Nm$

Therefore, the x component of the torque is 9.4 Nm, which corresponds to choice D.

Question 19

A square loop (L = 0.20 m) consists of 50 closely wrapped turns, each carrying a current of 0.50 A. The loop is oriented as shown in a uniform magnetic field of 0.40 T directed in the positive y direction. What is the magnitude of the torque on the loop?

Accepted Answer

The magnitude of the torque on a current loop can be calculated using the formula:

τ = IABsinθ

where τ is the torque, I is the current, A is the area of the loop, B is the magnetic field, and θ is the angle between the magnetic field and the normal to the loop.

In this case, the loop is a square with sides of 0.20 m. The area of the loop is A = L^2 = 0.04 m^2. The current is I = 0.50 A. The magnetic field is B = 0.40 T, directed in the positive y direction.

The angle θ between the magnetic field and the normal to the loop is zero, since the loop is oriented parallel to the magnetic field.

Plugging in the values, we get:

τ = (0.50 A)(0.04 m^2)(0.40 T)(sin 0)

τ = 0 N ⋅ m

However, this answer doesn't make sense, since we expect the loop to experience a torque in the presence of a magnetic field. The problem is that we need to take into account the fact that the loop consists of multiple turns of wire. Each turn of the loop contributes a torque to the total torque on the loop.

The torque on one turn can be calculated using the same formula:

τ = IABsinθ

where now A is the area of one turn of the loop. Since there are 50 turns in the loop, the total torque is:

τ = (50)(0.50 A)(0.002 m^2)(0.40 T)(sin 0)

τ = 0.35 N ⋅ m

Therefore, the magnitude of the torque on the loop is 0.35 N ⋅ m, and the best choice is (C).

Question 20

The figure shows the orientation of a rectangular loop consisting of 80 closely wrapped turns each carrying a current I. The magnetic field in the region is (  ) mT. The loop can turn about the y axis. If θ = 30°, a = 0.40 m, b = 0.30 m, and I = 8.0 A, what is the magnitude of the torque exerted on the loop?

Accepted Answer

The answer of The figure shows the orientation of a...

Quiz 29: Magnetic Fields

A straight 10-cm wire bent at its midpoint so as to form an angle of 90° carries a current of 10 A. It lies in the xy plane in a region where the magnetic field is in the positive z direction and has a constant magnitude of 3.0 mT. What is the magnitude of the magnetic force on this wire?

A proton moves around a circular path (radius = 2.0 mm) in a uniform 0.25-T magnetic field. What total distance does this proton travel during a 1.0-s time interval? (m = 1.67 × 10−²⁷ kg, q = 1.6 × 10−¹⁹ C)

What current must be maintained in a square loop (50 cm on a side) to create a torque of 1.0 N ⋅ m about an axis through its center and parallel to one of its sides when a magnetic field of magnitude 70 mT is directed at 40° to the plane of the loop?

A 500-eV electron and a 300-eV electron trapped in a uniform magnetic field move in circular paths in a plane perpendicular to the magnetic field. What is the ratio of the radii of their orbits?

A wire (mass = 50 g, length = 40 cm) is suspended horizontally by two vertical wires which conduct a current I = 8.0 A, as shown in the figure. The magnetic field in the region is into the paper and has a magnitude of 60 mT. What is the tension in either wire?

A charged particle (m = 2.0 g, q = −50 μC) moves in a region of uniform field along a helical path (radius = 4.0 cm, pitch = 8.0 cm) as shown. What is the angle between the velocity of the particle and the magnetic field?

A uniform magnetic field of 0.50 T is directed along the positive x axis. A proton moving with a speed of 60 km/s enters this field. The helical path followed by the proton shown has a pitch of 5.0 mm. Determine the angle between the magnetic field and the velocity of the proton.

A deuteron is accelerated from rest through a 10-kV potential difference and then moves perpendicularly to a uniform magnetic field with B = 1.6 T. What is the radius of the resulting circular path? (deuteron: m = 3.3 × 10−²⁷ kg, q = 1.6 × 10−¹⁹ C)

A rectangular coil (0.20 m × 0.80 m) has 200 turns and is in a uniform magnetic field of 0.30 T. When the orientation of the coil is varied through all possible positions, the maximum torque on the coil by magnetic forces is 0.080 N ⋅ m. What is the current in the coil?

A charged particle moves in a region of uniform magnetic field along a helical path (radius = 5.0 cm, pitch = 12 cm, period = 5.0 ms) . What is the speed of this particle as it moves along this path?

What is the magnetic force on a 2.0-m length of (straight) wire carrying a current of 30 A in a region where a uniform magnetic field has a magnitude of 55 mT and is directed at an angle of 20° away from the wire?

The boundary shown is that of a uniform magnetic field directed in the positive z direction. An electron enters the magnetic field with a velocity pointing along the x axis and exits 0.63 μs later at point A. What is the magnitude of the magnetic field?

A particle (m = 3.0 μg, q = 5.0 μC) moves in a uniform magnetic field given by ( ) mT. At t = 0 the velocity of the particle is equal to ( ) m/s. The subsequent path of the particle is

A circular loop (radius = 0.50 m) carries a current of 3.0 A and has unit normal vector of ( ) /3. What is the x component of the torque on this loop when it is placed in a uniform magnetic field of ( ) T?

A square loop (L = 0.20 m) consists of 50 closely wrapped turns, each carrying a current of 0.50 A. The loop is oriented as shown in a uniform magnetic field of 0.40 T directed in the positive y direction. What is the magnitude of the torque on the loop?

Quiz 29: Magnetic Fields

A straight 10-cm wire bent at its midpoint so as to form an angle of 90° carries a current of 10 A. It lies in the xy plane in a region where the magnetic field is in the positive z direction and has a constant magnitude of 3.0 mT. What is the magnitude of the magnetic force on this wire?

A proton moves around a circular path (radius = 2.0 mm) in a uniform 0.25-T magnetic field. What total distance does this proton travel during a 1.0-s time interval? (m = 1.67 × 10−27 kg, q = 1.6 × 10−19 C)

What current must be maintained in a square loop (50 cm on a side) to create a torque of 1.0 N ⋅ m about an axis through its center and parallel to one of its sides when a magnetic field of magnitude 70 mT is directed at 40° to the plane of the loop?

A 500-eV electron and a 300-eV electron trapped in a uniform magnetic field move in circular paths in a plane perpendicular to the magnetic field. What is the ratio of the radii of their orbits?

A wire (mass = 50 g, length = 40 cm) is suspended horizontally by two vertical wires which conduct a current I = 8.0 A, as shown in the figure. The magnetic field in the region is into the paper and has a magnitude of 60 mT. What is the tension in either wire?

A charged particle (m = 2.0 g, q = −50 μC) moves in a region of uniform field along a helical path (radius = 4.0 cm, pitch = 8.0 cm) as shown. What is the angle between the velocity of the particle and the magnetic field?

A uniform magnetic field of 0.50 T is directed along the positive x axis. A proton moving with a speed of 60 km/s enters this field. The helical path followed by the proton shown has a pitch of 5.0 mm. Determine the angle between the magnetic field and the velocity of the proton.

A deuteron is accelerated from rest through a 10-kV potential difference and then moves perpendicularly to a uniform magnetic field with B = 1.6 T. What is the radius of the resulting circular path? (deuteron: m = 3.3 × 10−27 kg, q = 1.6 × 10−19 C)

A rectangular coil (0.20 m × 0.80 m) has 200 turns and is in a uniform magnetic field of 0.30 T. When the orientation of the coil is varied through all possible positions, the maximum torque on the coil by magnetic forces is 0.080 N ⋅ m. What is the current in the coil?

A charged particle moves in a region of uniform magnetic field along a helical path (radius = 5.0 cm, pitch = 12 cm, period = 5.0 ms) . What is the speed of this particle as it moves along this path?

What is the magnetic force on a 2.0-m length of (straight) wire carrying a current of 30 A in a region where a uniform magnetic field has a magnitude of 55 mT and is directed at an angle of 20° away from the wire?

The boundary shown is that of a uniform magnetic field directed in the positive z direction. An electron enters the magnetic field with a velocity pointing along the x axis and exits 0.63 μs later at point A. What is the magnitude of the magnetic field?

A particle (m = 3.0 μg, q = 5.0 μC) moves in a uniform magnetic field given by ( ) mT. At t = 0 the velocity of the particle is equal to ( ) m/s. The subsequent path of the particle is

A circular loop (radius = 0.50 m) carries a current of 3.0 A and has unit normal vector of ( ) /3. What is the x component of the torque on this loop when it is placed in a uniform magnetic field of ( ) T?

A square loop (L = 0.20 m) consists of 50 closely wrapped turns, each carrying a current of 0.50 A. The loop is oriented as shown in a uniform magnetic field of 0.40 T directed in the positive y direction. What is the magnitude of the torque on the loop?

A proton moves around a circular path (radius = 2.0 mm) in a uniform 0.25-T magnetic field. What total distance does this proton travel during a 1.0-s time interval? (m = 1.67 × 10−²⁷ kg, q = 1.6 × 10−¹⁹ C)

A deuteron is accelerated from rest through a 10-kV potential difference and then moves perpendicularly to a uniform magnetic field with B = 1.6 T. What is the radius of the resulting circular path? (deuteron: m = 3.3 × 10−²⁷ kg, q = 1.6 × 10−¹⁹ C)