In order to solve this problem we are going to use the formula that defines the force acting on the current-carrying conductor by a magnetic field

$$
F=ILB
$$

From here, we can express the wire’s length as

$$
L=frac{F}{IB}
$$

If we now plug in the values we obtain the following

$$
L=frac{2.1}{7.2times 0.0089}
$$

Finally

$$
boxed{L=3.3 times 10^1textrm{m}}
$$

So, the answer is D.

$$
textrm{D; }L=3.3 times 10^1textrm{m}
$$

In order to solve this problem we are going to use the formula that defines the force acting on the current-carrying conductor by a magnetic field

$$
F=ILB
$$

From here, we can express the current as

$$
I=frac{F}{LB}
$$

If we now plug in the values we obtain the following

$$
I=frac{7.6times 10^{-3}}{0.19times 4.1}
$$

Finally

$$
boxed{I=9.8times 10^{-3}textrm{ A}}
$$

So, the answer is B.

$$
textrm{B; }I=9.8times 10^{-3}textrm{ A}
$$

In order to solve this problem we are going to use the formula that defines the force acting on a charged particle by a magnetic field

$$
vec F=qvec vtimes vec B=qvB
$$

If we now plug in the values we obtain the following

$$
F=7.12times 10^{-6}times 3times 10^8times 4.02times 10^{-3}
$$

Finally

$$
boxed{F=8.59textrm{ N}}
$$

So the answer is A.

$$
textrm{A; }F=8.59textrm{ N}
$$

In order to solve this problem we are going to use the formula that defines the force acting on a charged particle by a magnetic field

$$
vec F=qvec vtimes vec B=qvB
$$

From here, we can express the magnetic field as

$$
B=frac{F}{qv}
$$

If we now plug in the values we obtain the following

$$
B=frac{18}{1.6times 10^{-19}times 7.4times 10^5}
$$

Finally
$boxed{B=1.52 times 10^{14}textrm{ T}}$
So the answer is D.

$$
textrm{D; }B=1.52 times 10^{14}textrm{ T}
$$

To general formula of the magnetic field of a solenoid is given as
$B=mu_0mu_rfrac{NI}{L}$
so we see that it depends on the core type ($mu_r$), number of turns ($N$), and the strength of the current ($I$) but doesn’t depend on the thickness of the wire $d$. So the answer is C.

In order to solve this problem, we are going to use the formula that defines the force acting on a charged particle by a magnetic field

$$
vec F=qvec vtimes vec B=qvB
$$

If we now plug in the values we obtain the following

$$
F=1.6times 10^{-19}times 4times 10^6times 0.25
$$

Finally
$boxed{F=1.6times 10^{-13}textrm{ N}}$
And by using the first right-hand rule we obtain that the force points to the left so the answer is A.

$$
textrm{A) }F=1.6times 10^{-13}textrm{ N}
$$

To derive the units in two cases we are going to use unit definition of each physical property in the given formulas. In the first case

$$
F=qvB
$$

So we can express the magnetic field as

$$
B=frac{F}{qv}=frac{ma}{qv}
$$

Or in terms of units

$$
textrm{T}=frac{textrm{kg}cdotfrac{textrm{m}}{textrm{s}^2}}{textrm{C}cdot frac{textrm{m}}{textrm{s}}}
$$

$$
boxed{textrm{T}=frac{textrm{kg}}{textrm{C}cdottextrm{s}}}
$$

On the other hand, if we start from

$$
F=ILB
$$

$$
B=frac{F}{IL}=frac{ma}{frac{q}{t}L}
$$

Again, in terms of units

$$
T=frac{textrm{kg}cdotfrac{textrm{m}}{textrm{s}^2}}{{textrm{m}cdot frac{textrm{C}}{textrm{s}}}}
$$

which again gives

$$
boxed{textrm{T}=frac{textrm{kg}}{textrm{C}cdottextrm{s}} }
$$

$$
textrm{T}=frac{textrm{kg}}{textrm{C}cdottextrm{s}}
$$

In order to solve this problem, we have to find the current in the given circuit. To do so, we employ Ohm’s law which says

$$
I=frac{V}{R}=frac{5.8}{18}
$$

which gives that

$$
I=0.32textrm{A}
$$

Now, the angle can be found from the formula

$$
F=ILBsintheta
$$

which can be solved for $theta$ to give

$$
theta=arcsin(frac{F}{ILB})
$$

$$
theta=arcsin(frac{22times 10^{-3}}{0.32times 0.14times 0.85})
$$

Which finally gives us that

$$
boxed{theta=35.3^circ}
$$

$$
theta=35.3^circ
$$