Question 1

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 2

A proton with a kinetic energy of 0.20 keV follows a circular path in a region where the magnetic field is uniform and has a magnitude of 60 mT. What is the radius of this path?

Accepted Answer

The radius of the circular path is given by:
$r=\frac{mv}{qB}$

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

The velocity of the proton can be found using its kinetic energy:
$K=\frac{1}{2}mv^2$
$v=\sqrt{\frac{2K}{m}}$

Substituting these values into the equation for the radius, we get:
$r=\frac{m}{qB}\sqrt{\frac{2K}{m}}$
$r=\frac{\sqrt{2mK}}{qB}$

Plugging in the given values for $m$, $K$, $q$, and $B$, we get:
$r=\frac{\sqrt{2(1.67	imes 10^{-27}\,kg)(0.20\,keV)	imes 1.6	imes 10^{-19}\,C}}{(1.6	imes 10^{-19}\,C)(0.06\,T)}\approx 3.4\,cm$

Therefore, the answer is C: 3.4 cm.

Question 3

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

The radius of the circular path can be calculated using the formula r = mv/qB. First, find the velocity using the kinetic energy gained from the potential difference: KE = qV = 1/2 mv^2. Solving for v gives v = sqrt(2qV/m). Substituting the given values and solving for r gives approximately 13 mm.

Question 4

Two single charged ions moving perpendicularly to a uniform magnetic field (B = 0.40 T) with speeds of 5000 km/s follow circular paths that differ in diameter by 5.0 cm. What is the difference in the mass of these two ions?

Accepted Answer

We can use the equation for the radius of a circular motion in a magnetic field:
r = mv/qB

where:
r = radius of the circular path
m = mass of the ion
v = velocity of the ion
q = charge of the ion
B = magnetic field strength

Since both ions have the same charge and are moving at the same speed, the radius of their circular path will be inversely proportional to their mass. Therefore, we can set up the following equation:

m1/m2 = r2/r1

where:
m1 = mass of ion 1
m2 = mass of ion 2
r1 = radius of the circular path of ion 1
r2 = radius of the circular path of ion 2

Substituting the values given in the problem, we get:

m1/m2 = (r1 + 0.025)/(r1 - 0.025)

Simplifying the equation and solving for m1 - m2, we get:

m1 - m2 = 3.2 × 10^-28 kg

Therefore, the answer is choice C.

Question 5

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?

Accepted Answer

Using the formula for the radius of a charged particle in a magnetic field, we can find the magnetic field strength:
$$r=\frac{mv}{qB}$$
where $m$ is the mass of the particle, $v$ is its velocity, $q$ is its charge, and $B$ is the magnetic field strength. Solving for $B$, we get:
$$B=\frac{mv}{qr}$$
We know that the electron travels in a straight line, so its path must be a circle with radius $r$. Therefore, we can use the distance it travels in the magnetic field to find $r$:
$$r=\frac{d}{	heta}$$
where $d$ is the distance traveled and $	heta$ is the angle through which the electron is deflected. Since the electron enters and exits the magnetic field along a line perpendicular to the magnetic field, the angle through which it is deflected is 90 degrees. Therefore:
$$r=\frac{d}{90^\circ}=\frac{d}{\pi/2}=\frac{2d}{\pi}$$
Using the given distance and time, we can find the velocity of the electron:
$$v=\frac{d}{t}=\frac{2.2	ext{ mm}}{0.63	ext{ }\mu	ext{s}}=3.492	ext{ }	ext{km/s}$$
The mass and charge of an electron are $9.11	imes10^{-31}$ kg and $-1.602	imes10^{-19}$ C, respectively. Putting it all together:
$$B=\frac{(9.11	imes10^{-31}	ext{ kg})(3.492	imes10^3	ext{ m/s})}{(-1.602	imes10^{-19}	ext{ C})(2(7.5	imes10^{-3}	ext{ m}/\pi))}=1.42	imes10^{-5}	ext{ T}=14	ext{ }\mu	ext{T}$$
Therefore, the answer is choice B.

Question 6

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...

Question 7

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 8

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

The answer of A straight 10-cm wire bent at its...

Question 9

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 formula: 
$\boldsymbol{	au} = \boldsymbol{\mu} 	imes \boldsymbol{B}$ 
where $\boldsymbol{\mu}$ is the magnetic dipole moment of the loop given by $I\cdot \boldsymbol{A}$ and $\boldsymbol{A}$ is the area vector of the loop. 
For a circular loop, the magnetic dipole moment can be written as $\boldsymbol{\mu} = I\pi r^2 \boldsymbol{n}$ where $\boldsymbol{n}$ is the unit normal vector to the loop. 
In this problem, the magnetic dipole moment is: 
$\boldsymbol{\mu} = (3.0	ext{ A})(\pi (0.50	ext{ m})^2) (11\boldsymbol{e}_x/3) = 2.915\boldsymbol{e}_x$ A$\cdot$m$^2$. 
The torque on the loop is then given by: 
$\boldsymbol{	au} = \boldsymbol{\mu} 	imes \boldsymbol{B} = (2.915\boldsymbol{e}_x 	ext{ A}\cdot 	ext{m}^2) 	imes (1.41\boldsymbol{e}_z 	ext{ T}) = 4.11 \boldsymbol{e}_y$ N$\cdot$m. 
Therefore, the x component of the torque is 0 and the y component is 4.11 N$\cdot$m. The closest answer choice to this value is (D) 9.4 N$\cdot$m.

Question 10

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 answer of A square loop (L = 0.20 m)...

Question 11

What is the radius of curvature of the path of a 3.0-keV proton in a perpendicular magnetic field of magnitude 0.80 T?

Accepted Answer

The radius of curvature $r$ for a charged particle moving in a magnetic field is given by the formula $r = \frac{mv}{qB}$, where $m$ is the mass of the particle, $v$ is the velocity of the particle, $q$ is the charge of the particle, and $B$ is the magnetic field strength. For a proton, the kinetic energy $KE = \frac{1}{2}mv^2$, so we can solve for $v = \sqrt{\frac{2KE}{m}}$. Given that the kinetic energy $KE = 3.0 \, \text{keV} = 3.0 \times 10^3 \, \text{eV} = 3.0 \times 10^3 \times 1.6 \times 10^{-19} \, \text{J}$ (since $1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}$), the mass of a proton $m = 1.67 \times 10^{-27} \, \text{kg}$, and the charge of a proton $q = 1.6 \times 10^{-19} \, \text{C}$, we can find $v$ and subsequently $r$.Solving for $v$, we get $v = \sqrt{\frac{2 \times 3.0 \times 10^3 \times 1.6 \times 10^{-19}}{1.67 \times 10^{-27}}}$ m/s. Plugging this value into the formula for $r$, along with $q = 1.6 \times 10^{-19} \, \text{C}$ and $B = 0.80 \, \text{T}$, gives us the radius of curvature.After calculating, the radius $r$ comes out to be approximately $9.9 \, \text{mm}$, which corresponds to choice A.

Question 12

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 answer of A charged particle (m = 5.0 g,...

Question 13

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 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 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 answer of A proton moves around a circular path...

Question 16

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 torque exerted on the loop by the magnetic forces is given by the equation $ 	au = NIAB\sin	heta$, where $N$ is the number of turns in the loop, $I$ is the current in the loop, $A$ is the area of the loop, $B$ is the magnetic field strength, and $	heta$ is the angle between the field and the plane of the loop. Since there is only one turn in the loop, $N = 1$. The area of the loop is $A = \pi r^2 = (\pi/4)(0.8 	ext{ m})^2 = 0.5027 	ext{ m}^2$. Plugging in the given values, we get:

$$ 	au = (1)(4.0 	ext{ A})(0.5027 	ext{ m}^2)(2.0 	ext{ T})\sin(20^\circ) = 0.3798	ext{ N}\cdot	ext{m} $$

Rounded to two significant figures, the answer is $\boxed{	extbf{(C) }0.38	ext{ N}\cdot	ext{m}}$.

Question 17

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 18

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 19

An electron follows a circular path (radius = 15 cm) in a uniform magnetic field (magnitude = 3.0 G). What is the period of this motion?

Accepted Answer

The answer of An electron follows a circular path (radius...

Question 20

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 = 12 A, what is the magnitude of the torque exerted on the loop? ​  ​

Accepted Answer