Solving Problems Involving Magnetic Forces
- There is a lot of mathematics surrounding magnetism and magnetic forces
- The key equations are:
Force on a current-carrying conductor:
F = BIL
F = BIL
Force on a moving charge:
F = Bqv
F = Bqv
Radius of a moving charge in a magnetic field:
r =
r =
- Below are two worked examples demonstrating different situations involving magnetic forces
Worked example
A 5 cm length of wire is at 90° to the direction of an external magnetic field. When a current of 1.5 A flows through the wire it experiences a force of 0.06 N from the motor effect.
Calculate the magnetic flux density of the magnet.
Calculate the magnetic flux density of the magnet.
Step 1: List the known quantities
- Length, L = 5 cm = 0.05 m
- Current, I = 1.5 A
- Force, F = 0.06 N
Step 2: Write out the equation for magnetic force
F = BIL
Step 3: Rearrange the equation to make B the subject
Step 4: Substitute values into the equation
Worked example
This question is about the movement of an electrically charged particle into a magnetic field. An electron enters a magnetic field and moves in an approximately circular path as shown below.
- Explain whether the speed of the proton is the same when entering and exiting the magnetic field.
- The magnetic field has a strength of 0.003 T and the velocity of the electron before entering the magnetic field is 8.6 × 106 m s−1 to the left. Show that the radius of the motion of the electrons is 1.63 cm.
Part (a)
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The answer is that the work done by the magnetic force on the charge must be zero
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This is because the force itself is at right angles to the velocity
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Since the work done is zero, therefore the kinetic energy does not change between entering and leaving the magnetic field
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Therefore the speed is the same
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Part (b)
Step 1: List the known quantities
- Magnetic field strength, B = 0.003 T
- Velocity of electron (before entering field), v = 8.6 × 106 m s−1 to the left
- Radius of motion to be shown, r = 1.63 cm
Step 2: Equate the magnetic force and the force of circular motion
- This can be done for the situation since the circular motion is caused by the magnetic force
Step 3: Rearrange the equation to find the radius
qvBr = mv2
qBr = mv
Step 4: Substitute in the values
Step 5: State final answer
R = 1.63 × 10−2 m = 1.63 cm in a circular motion