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 can set up a circuit with a power supply connected to an aluminum foil strip placed in a uniform magnetic field produced by a magnet.

1. Connect the aluminum foil to a power supply to create a current.
2. Position the foil strip perpendicular to the magnetic field lines.
3. Observe the deflection of the foil strip, which indicates a force acting on it due to the magnetic field.

A current-carrying conductor will not experience a force if it is parallel to the magnetic field lines. In this position, the angle between the current direction and the magnetic field is 0 degrees, resulting in no force acting on the conductor.

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

$F = B imes I imes l$
Where:

• F is the force,
• B is the magnetic flux density,
• I is the current, and
• l is the length of the wire.

Use the provided data to plot a graph with current (I) on the x-axis and force (F) on the y-axis, ensuring to label the axes consistently.

The slope of the graph (m) represents the ratio of change in force to change in current. Using the points (0.5 A, 10 mN) and (3.5 A, 68 mN):

ext{slope} = rac{(68 - 10) ext{ mN}}{(3.5 - 0.5) ext{ A}} = rac{58 ext{ mN}}{3.0 ext{ A}} = 19.33 ext{ mN/A}

Substituting into the expression we found earlier:
B = rac{F}{I imes l} = rac{19.33 ext{ mN}}{(0.03 ext{ m})} = 0.6444 ext{ T}

Starting with the concept of force acting on a moving charge, we can use the definition of magnetic force:

1. The force F on a charge q moving at velocity v in a magnetic field B that is perpendicular to its motion is defined by:
   $F = qvB$
   This shows how the force depends on the charge, its speed, and the strength of the magnetic field.

Given:

• Magnetic flux density B = 0.5 T
• Radius of circular path r = 2.3 m
  The centripetal force acting on the proton can be given by:
  F = rac{mv^2}{r}
  This force is also equal to the magnetic force:
  $F = qvB$
  Thus,
  rac{mv^2}{r} = qvB
  Rearranging gives:
  v = rac{qBr}{m}
  Using the charge of a proton $q = 1.6 imes 10^{-19} ext{ C}$ and mass $m = 1.67 imes 10^{-27} ext{ kg}$:

v = rac{(1.6 imes 10^{-19} ext{ C}) (0.5 ext{ T}) (2.3 ext{ m})}{(1.67 imes 10^{-27} ext{ kg})} = 1.11 imes 10^7 ext{ m/s}
This is the calculated speed of the proton as it enters the field.