Home Page Textbooks Physics Page 815: Standardized Test Prep

Physics

1st Edition

Walker

ISBN: 9780133256925

Textbook solutions

All Solutions

Page 815: Standardized Test Prep

Exercise 1

Step 1

1 of 4

The magnitude of the magnetic force of a moving charge is calculated using
$$
begin{aligned}
F = lvert q rvert vB sin theta
end{aligned}
$$
where $|q|$ is the magnitude of the charge, $v$ is the velocity of the object, $B$ is magnefic field, and $theta$ is the angle between the direction of the object and the magnetic field.

**GIVEN**

Charge of electron: $q = 1.6 times 10^{-19};text{C}$
Magnetic field: $B = 200;text{T}$
Velocity: $v = 2.0times 10^{5};text{m/s}$
Angle: $theta = 90degree$

Step 2

2 of 4

We substitute the known variables into the equation for the magnetic force.
$$
begin{aligned}
F = (1.6 times 10^{-19};text{C})(2.0times 10^{5};text{m/s})(B = 200;text{T}) vB sin 90degree = boxed{6.4times 10^{-12};text{N}}
end{aligned}
$$

Step 3

3 of 4

By the virtue of right hand rule, the direction of the magnetic force is into the page. However, we are considering an electron with a negative charge. Hence, the direction of magnetic force is the opposite of the right hand rule. Thus, **the magnetic force is out of the page.**

Result

4 of 4

Exercise 2

Step 1

1 of 4

The magnitude of the magnetic force is dependent on the angle $theta$ between the velocity and magnetic field as shown by the equation:
$$begin{align*}
F=vert q vert vBsin theta
end{align*}$$

Step 2

2 of 4

Changing the direction of the velocity of the electron to a certain angle will therefore change the angle $theta$ between the velocity and the magnetic field which would result in a change in the magnitude of the magnetic force.

Step 3

3 of 4

However, the $textbf{direction of the magnetic force}$ $textbf{would still
be the same as before.}$ By the right hand rule, the direction of the magnetic force would still into the page

Result

4 of 4

$text{B.}$

Exercise 3

Step 1

1 of 4

The magnitude of the magnetic force is shown by the equation:
$$begin{align*}
F=vert q vert vBsin{theta}
end{align*}$$
where $theta$ is the angle between the velocity vector $vec{v}$ and magnetic field vector $vec{B}$.

Step 2

2 of 4

If a proton with charge $q$ moves along the same direction as the magnetic field, the angle between them would then be $0^{circ}$. Plugging in $theta$ would result to:
$$begin{align*}
F=vert q vert vBsin(0^{circ})=vert q vert vB (0)= 0;text{N}
end{align*}$$

Step 3

3 of 4

Therefore, $textbf{the force is zero}$ when the proton moves in the same direction as the magnetic field line.

Result

4 of 4

$text{D.}$

Exercise 4

Step 1

1 of 7

The magnitude of the electric force due to an electric field is given by:
$$begin{align}
F=vert q vert E
end{align}$$
The magnitude of the magnetic force due to a magnetic field is given by:
$$begin{align}
F=vert q vert vB sin{theta}
end{align}$$

Step 2

2 of 7

$textbf{(a)}$

The electric force is not dependent on the velocity $v$ of the charge. To solve for this force, we use Equation 1 with given quantities: $E=400frac{text{N}}{text{m}}$ and $q=1.60times10^{-19};text{C}$
$$begin{align*}
F=vert (1.60times10^{-19};text{C}) vert (400frac{text{N}}{text{m}})=64times10^{-18};text{N}
end{align*}$$

Step 3

3 of 7

$textbf{(b)}$

As stated earlier, the velocity $v$ of the charge does not matter in calculating the electric force. Thus, we can easily get the force through Equation 1, given $E=800frac{text{N}}{text{m}}$ and $q=-1.60times10^{-19};text{C}$.
$$begin{align*}
F=vert (-1.60times10^{-19};text{C}) vert (800frac{text{N}}{text{m}})=128times10^{-18};text{N}
end{align*}$$

Step 4

4 of 7

Note that, in contrast with the electric force, the magnetic force takes into consideration the velocity $v$ of the charge. Using Equation 2, we calculate the magnetic force with given quantities: $60000frac{text{m}}{text{s}}$, $B=800;text{T}$ and $q=1.60times10^{-19};text{C}$.
$$begin{align*}
F &=vert (1.60times10^{-19};text{C}) vert (60000frac{text{m}}{text{s}})(800;text{T}) sin{90^{circ}}\
&=vert (1.60times10^{-19};text{C}) vert (60000frac{text{m}}{text{s}})(800;text{T})\
&=7.68times10^{-12};text{N}
end{align*}$$

Step 5

5 of 7

The proton is said to be stationary meaning $v=0frac{text{m}}{text{s}}$. As seen in Equation 2, the magnetic force is dependent on the velocity of the charge. Thus,
$$begin{align*}
F=0;text{N}
end{align*}$$

Step 6

6 of 7

From the calculations, the greatest force exerted on a charged particle is $7.68times10^{-12};text{N}$ which is the force in $textbf{C}$.

Result

7 of 7

$text{C.}$

Exercise 5

Step 1

1 of 1

The magnetic field right-hand rule for a current carrying wire states that the thumb points to the direction of the current $I$ and the fingers curl in the direction of the magnetic field $vec{B}$. With this rule, a current going to the right will have a magnetic field going $textbf{out of the page}$ at point P due to the top wire.

Exercise 6

Step 1

1 of 2

Force per unit length is given by

$$
frac{F}{Delta L}=frac{mu_{0}I_{1}I_{2}}{2pi r}
$$

Here we have $I_{1}=I_{2}=0.5$ A and $r=0.2$ cm $=0.002$ m. Hence
the force per meter is

$$
frac{F}{Delta L}=frac{left(1.26times10^{-6} {rm kgms^{-2}A^{-2}}right)left(0.5 {rm A}right)left(0.5 {rm A}right)}{2pileft(0.002 {rm m}right)}=2.5times10^{-5} {rm N/m}
$$

And since the current are flowing in the same direction the force
is attractive and hence force on bottom wire is upward direction.

Result

2 of 2

Correct option is (C).

Exercise 7

Step 1

1 of 6

We let $B_1$ be the magnetic field flowing at wire 1 and $B_2$ be the magnetic field flowing at wire 2. From the given diagram, it can be inferred that the total magnetic field experienced at point P is the sum of $B_1$ and $B_2$ because the two magnetic field flows on the same direction.

The magnetic field produced by a long, straight wire is given by
$$begin{align}
B = frac{mu_0 I}{2pi r}
end{align}$$
where the permeability of free space is $mu_0 = 4pitimes 10^{-7};text{T}cdot text{m/A}$, $I$ is the current running through it, and $r$ is the radial distance from wire.

Step 2

2 of 6

$textbf{(a)}$

We first find $B_1$ by plugging in the following values on Equation 1: $I = 0.5;text{A}$ and $r = 0.002;text{m}$.
$$begin{align*}
B_1 = frac{left(4pitimes 10^{-7};text{T}cdot frac{text{m}}{text{A}}right) (0.5;text{A})}{2pi (0.002;text{m})} = 5times 10^{-5};text{T}
end{align*}$$

Step 3

3 of 6

We find $B_2$ by plugging in the following values on Equation 1: $I = 0.5;text{A}$ and $r = 0.004;text{m}$.
$$begin{align*}
B_1 = frac{left(4pitimes 10^{-7};text{T}cdot frac{text{m}}{text{A}}right) (0.5;text{A})}{2pi (0.004;text{m})} = 2.5times 10^{-5};text{T}
end{align*}$$

Step 4

4 of 6

We then add $B_1$ and $B_2$ to find the total magnetic field experienced at point P.
$$begin{align*}
B_{text{P}} &= B_1 + B_2 \ &= 5times 10^{-5};text{T} + 2.5times 10^{-5};text{T} \ &= boxed{7.5times 10^{-5};text{T}}
end{align*}$$

Step 5

5 of 6

$textbf{(b)}$

Since the top wire is reversed, the magnetic field produced by two wires do not flow on the same direction. Hence, the magnetic field experienced at point P is the sum of $-B_1$ (into the page)and $B_2$ (out of page).
$$begin{align*}
B_{text{P}}’ &= B_1 + B_2 \ &= -5times 10^{-5};text{T} + 2.5times 10^{-5};text{T} \ &= boxed{-2.5times 10^{-5};text{T} text{ or } 2.5times 10^{-5};text{T};text{into the page}}
end{align*}$$

Result

6 of 6

(a) $7.5times 10^{-5};text{T}$, out of the page
(b) $2.5times 10^{-5};text{T}$, into the page

unlock

All Solutions

Page 815: Standardized Test Prep

Haven't found what you were looking for?

Search for samples, answers to your questions and flashcards