Both a moving charge and a conductor carrying an electric current experience a force in a magnetic field - Leaving Cert Physics - Question 9 - 2019

To demonstrate this, one could set up an experiment using a power supply, aluminium foil as the conductor, and a pair of magnets.

1. Connect the aluminium foil to the power supply to create a circuit.
2. Place the magnets so that their magnetic field is perpendicular to the current flowing through the aluminium foil.
3. Observe the movement of the foil; it should move due to the force exerted on it by the magnetic field, confirming that a current-carrying conductor experiences a force.

A current-carrying conductor will not experience a force when it is aligned parallel to the magnetic field lines. In such cases, the angle between the current direction and the magnetic field is 0°, and thus the force is zero according to the formula:

$F = BIl ext{sin}( heta) \ ext{where } F = 0 \ ext{at } heta = 0°$

The expression for the force F on the current-carrying wire is given by:

$F = BIl$

where:

• F is the force in newtons (N)
• B is the magnetic flux density (T)
• I is the current (A)
• l is the length of the wire (m)

1. Label the axes with 'Current (I)' on the x-axis and 'Force (F)' on the y-axis.
2. Plot the points from the data provided.
3. Draw a best-fit line to represent the relationship.

The slope of the graph will give us the magnetic flux density B, calculated using:

$B = \frac{\Delta F}{\Delta I}$

Starting from the expression for the force on a current-carrying conductor, we know:

$F = BIl$

Where I can be expressed as:

$I = \frac{q}{t}$

Substituting this into the expression for force, we get:

$F = B \left( \frac{q}{t} \right) l$

Given that velocity $v = \frac{l}{t}$, we can rearrange to find t:

$t = \frac{l}{v}$

Now substituting into the equation gives:

$F = Bq \left( \frac{l}{l/v} \right) = qvB$

To calculate the speed of the proton, we use the given formula:

$F = qvB$

Where:

• F can be found from the circular motion of the proton: $F = \frac{mv^2}{r}$
• For a proton: $q = 1.6 \times 10^{-19} C$ and the radius $r = 2.3 m$, and magnetic flux density $B = 0.5 T$.

By equating these:

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

Rearranging gives:

$v = \frac{qBr}{m}$

Substituting values gives:

$v = \frac{(1.6 \times 10^{-19})(0.5)(2.3)}{(1.67 \times 10^{-27})} \Rightarrow v \approx 1.1 \times 10^5 m/s$