The equation for force on a moving charged particle in a magnetic field is

$\displaystyle F=qv\cdot B$.

Because the charge is moving perpendicularly through the magnetic field, we don't have to worry about the cross product, and the equation becomes simple multiplication. 

$\displaystyle F&=qvB$

$\displaystyle F= (11\mu C)(4.7\cdot 10^7\frac{m}{s})(7T)$

$\displaystyle F= 3619$

Therefore, the force experienced by the particle is 3619N.

For a conductor moving in a magnetic field, cutting straight across the field lines, a potential is generated $\displaystyle \varepsilon=Blv$ where $\displaystyle \varepsilon$ is the potential or EMF, $\displaystyle B$ is the magnetic field strength, and $\displaystyle v$ is the conductor's velocity relative to the field.

$\displaystyle \varepsilon=0.022T* 0.12m*17\frac{m}{s} =0.045V$

As the rod moves, the positive charge feels an upward-directed force by the right-hand rule, and the negative a downward force resulting in the top being at a higher potential than the bottom of the rod.

The separated charge creates a potential $\displaystyle \varepsilon=Blv$. This potential results in an electric field $\displaystyle E=\frac{\varepsilon}{l}=\frac{Blv}{l}=Bv$ When this induced electric field creates a force $\displaystyle F=qE$ equal to the magnetic force $\displaystyle F=qvB$ on the mobile charge carriers, motion stops. Of course, if an electric circuit is created drawing current from the rod, motion will resume to rebuild the field.

To solve this question, we need to relate the speed and charge of the particle with the magnetic force it experiences in order to solve for the magnetic field strength. Thus, we'll need to use the following equation:

$\displaystyle F_{B}=qvBsin(\theta )$

Also, we are told that the particle is moving perpendicularly to the magnetic field.

$\displaystyle \theta=90^{o}$

$\displaystyle \sin \left ( 90^{o}\right )=1$

$\displaystyle F_{B}=qvB$

Rearrange to solve for the magnetic field, then plug in known values and solve.

$\displaystyle B=\frac{F_{B}}{qv}$

$\displaystyle B=\frac{1.2*10^{-18} N}{\left ( 1.6* 10^{-19}C\right )(5* 10^{3}\frac{m}{s})}$

$\displaystyle B=1.5* 10^{-3}\frac{N\cdot s}{C\cdot m}$

To answer this question, we need to realize that the particle is moving in a circular path because of some sort of centripetal force. Since the charge is moving while within a constant magnetic field, we can conclude that it is the magnetic force that is responsible for the centripetal force that keeps this charge moving in a circle. Thus, we need to relate the centripetal force to the magnetic force.

$\displaystyle F_{c}=F_{B}$

$\displaystyle \frac{mv^{2}}{r}=qvB$

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

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

The above equation shows us that the radius of the circular path is directy proportional to the mass and velocity of the particle, and inversely proportional to the charge of the particle and the magnetic field strength. Thus, if the value of the magnetic field is doubled, the above equation predicts that the value of the radius would be cut in half.

Using:

$\displaystyle B=\mu_0\frac{N*I}{L}$

Where:

$\displaystyle B$ is the magnetic field

 $\displaystyle N$ is the number of coils

$\displaystyle I$ is the current in the solenoid

$\displaystyle L$ is the length of the solenoid

$\displaystyle \mu_0$ is $\displaystyle 1.257*10^{-6}\frac{N}{A^2}$

Plugging in values:

$\displaystyle B=1.257*10^{-6}\frac{600*3}{.600}$

$\displaystyle B=3.77*10^{-3}T$

Using the Biot-Savart law:

$\displaystyle B=\frac{\mu_0}{4\pi}*\frac{2\pi R^2 I}{(z^2+R^2)^{(3/2)}}$

Where $\displaystyle R$ is the radius of the loop

$\displaystyle I$ is the current

$\displaystyle z$ is the distance from the center of the loop

$\displaystyle \mu_0=4\pi*10^{-7}\frac{T\cdot m}{A}$

Plugging in values:

$\displaystyle B=\frac{4\pi*10^{-7}}{4\pi}*\frac{2\pi*.03^2*.250}{(0^2+.03^2)^{(3/2)}}$

$\displaystyle B=5.24*10^{-6}\textup{ T}$

Reversing the battery will reverse the direction of the current. Using the right hand rule, it can be seen that this will also reverse the direction of the magnetic field. Since the magnitude of the current stays the same, the magnitude of the magnetic field will as well.

Based on the Biot-Savart law:

$\displaystyle B=\frac{\mu_0}{4\pi}*\frac{2\pi R^2 I}{(z^2+R^2)^{(3/2)}}$

Doubling the voltage will double the current, which will double the magnetic field. The direction will stay the same.

Magnets will align themselves with the surrounding magnetic field. Thus, if the north pole of a magnet is pointing north, the direction of the magnetic field must be pointing north. Magnetic fields point towards magnetic south poles, so the geographic north pole is actually a magnetic south pole.