AP Physics 2 - Magnetic Fields

A circuit contains a battery and a $20\Omega$ resistor in series. Determine the magnitude of the magnetic field outside of the loop away from the wire.

None of these

Explanation

Using

$B=\frac{\mu_0*I}{2\pi*r}$

Converting to and plugging in values

$B=\frac{4\pi*10^{-7}*I}{2\pi*.002}$

Determining current:

$B=\frac{4\pi*10^{-7}*3}{2\pi*.002}$

There is a loop with a radius of $0.03\textup{ m}$ and a current of $0.250\textup{ A}$ . Determine the magnitude of the magnetic field at the center of the loop.

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

None of these

$B=7.35*10^{-6}\textup{ T}$

$B=8.44*10^{-6}\textup{ T}$

$B=6.11*10^{-6}\textup{ T}$

Explanation

Using the Biot-Savart law:

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

Where is the radius of the loop

is the current

is the distance from the center of the loop

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

Plugging in values:

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

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

If the north end of a magnetic points towards the geographic north pole, that means that the geographic north pole is a magnetic __________ pole.

South

North

Electrical

Mono

None of these

Explanation

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.

Radius of moon:

A loop of current carrying wire runs along the equator of the moon. Determine the magnetic field at the center of the loop if are traveling through it

$3.61*10^{-7}T$

None of these

$7.77*10^{-7}T$

$8.85*10^{-7}T$

$3.14*10^{-7}T$

Explanation

Using

$B=\frac{\mu_0 I}{2R}$

Plugging in values

$B=\frac{4\pi*10^{-7}* 1*10^6}{2*1.74*10^6}$

$B=3.61*10^{-7}T$

A circular circuit is powered by a $9\textup{ V}$ battery. How will the magnetic field change if the battery is removed and placed in the opposite direction?

The magnetic field will have the same magnitude, albeit in the opposite direction

None of these

The magnetic field will become zero

The magnetic field will have the same magnitude and direction

The magnetic field will double in magnitude and flip directions

Explanation

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.

How strong would a magnetic field need to be in order to make a particle with a mass of $4\cdot10^{-11}:kg$ and a charge of move in a circular path with a speed of $400:\frac{m}{s}$ and a radius of ?

$4\cdot10^{-9}:T$

$4\cdot10^{-6}:T$

$4\cdot10^{4}:T$

$4\cdot10^{2}:T$

Explanation

For this question, we are being asked to determine the magnetic field necessary to make a particle of a given mass and charge to move in a circular path with a given speed and radius.

To begin with, we can realize that the particle will be moving in a circular path. Thus, there is going to be a centripetal force associated with this circular motion. Moreover, because we know the particle will be present in a magnetic field, we can infer that the magnetic force will be the source of the centripetal force. Thus, we can start by writing out the expression for each of these forces, and then setting them equal to one another.

$F_{B}=qvB$

$F_{c}=\frac{mv^{2}}{r}$

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

Rearranging the above expression to isolate the term for magnetic field, we arrive at the following expression.

$B=\frac{mv^{2}}{qvr}=\frac{mv}{qr}$

Now, we can plug in the values given to us in the question stem to solve for the magnetic field strength.

$B=\frac{(4\cdot10^{-11}:kg)(400:\frac{m}{s})}{(8\cdot 10^{-9}:C)(0.5:m)}$

A circular circuit is powered by a $9\textup{ V}$ battery. How will the magnetic field change if a second $9\textup{ V}$ battery is added in the same direction as the first?

The magnetic field will double in magnitude and have the same direction.

None of these

The magnetic field will become zero

The magnetic field will quadruple

The magnetic field will stay the same

Explanation

Based on the Biot-Savart law:

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

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

An infinitely long wire carries a current of determine the magnitude of the magnetic field away.

$1.27*10^{-5}T$

None of these

$2.5*10^{-5}T$

$3.34*10^{-5}T$

$1.78*10^{-5}T$

Explanation

Magnetic field of an infinitely long wire:

$B=\frac{\mu_0*I}{2\pi *r}$

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

Plugging in values:

$B=\frac{4*10^{-7}*3}{2\pi *.015}$

$B=1.27*10^{-5}T$

Suppose that a positively charged particle with charge moves in a circular path of radius in a constant magnetic field of strength . If the magnetic field strength is doubled to , what effect does this have on the radius of the circular path that this charge takes?

$\frac{1}{2}r$

$\frac{1}{4}r$

Explanation

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.

$F_{c}=F_{B}$

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

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

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

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.

loops of current carrying wire form a solenoid of length $.600\textup{ m}$ that carries $3\textup{ A}$ and have radius $45\textup{ cm}$ . Determine the magnetic field at the center of the solenoid.