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CIE IGCSE Physics (0625) Force on a current-carrying conductor Study Notes

CIE IGCSE Physics (0625) Force on a current-carrying conductor Study Notes - New Syllabus

CIE IGCSE Physics (0625) Force on a current-carrying conductor Study Notes

LEARNING OBJECTIVE

  • Understanding the concepts of Force on a current-carrying conductor  

Key Concepts: 

  • Experiment: Force on a Current-Carrying Conductor in a Magnetic Field
  • Relationship Between Force, Magnetic Field, and Current

CIE iGCSE Physics (0625)  Study Notes – All topics

Experiment: Force on a Current-Carrying Conductor in a Magnetic Field

Experiment: Force on a Current-Carrying Conductor in a Magnetic Field

Objective: To demonstrate that a force acts on a wire carrying a current placed in a magnetic field, and to observe how changing the direction of the current or magnetic field affects the direction of the force.

Apparatus:

  • Strong horseshoe magnet
  • Straight conducting wire (e.g., copper)
  • Variable power supply
  • Switch
  • Connecting wires
  • Clamp stand to suspend the wire horizontally

Procedure:

  • Set up the wire horizontally between the poles of the horseshoe magnet so that the magnetic field is perpendicular to the wire (i.e., into or out of the page).
  • Connect the wire to a power supply with a switch so that you can control the direction and magnitude of the current.
  • Close the switch to allow current to flow through the wire.

 Observation:

When the current flows through the wire, the wire experiences a sideways push — this is the magnetic force acting on the current-carrying conductor.

Reversing the Current:

  • Reverse the connections to the power supply.
  • Now current flows in the opposite direction through the wire.
  • Observation: The wire is pushed in the opposite direction compared to before.

Reversing the Magnetic Field:

  • Flip the magnet so that the north and south poles are swapped.
  • Keep the direction of the current the same.
  • Observation: Again, the wire moves in the opposite direction compared to the original setup.

 Conclusion:

A current-carrying conductor placed in a magnetic field experiences a force. The direction of this force depends on both:

  • The direction of the current
  • The direction of the magnetic field

This phenomenon is described by Fleming’s Left-Hand Rule:

Thumb = Force, First finger = Field, Second finger = Current

Relationship Between Force, Magnetic Field, and Current

Relationship Between Force, Magnetic Field, and Current

Fleming’s Left-Hand Rule:

Fleming’s Left-Hand Rule is used to determine the direction of the magnetic force on a current-carrying conductor or a moving charge in a magnetic field.

Hold your left hand with the:

  • Thumb → Force (motion of conductor or particle)
  • First Finger → Magnetic Field (from North to South)
  • Second Finger → Current (conventional current: positive to negative)

 For Beams of Charged Particles

Charged particles (like electrons or protons) moving through a magnetic field also experience a force.

The direction of this force depends on:

  • Direction of the particle’s velocity
  • Direction of the magnetic field
  • Sign of the charge on the particle

Rule: Use Fleming’s Left-Hand Rule for positive charges (like protons). For negative charges (like electrons), the force is in the opposite direction to what the rule gives.

Example:

A proton is moving to the right. It enters a uniform magnetic field pointing into the page. What is the direction of the magnetic force on it?

▶️ Answer/Explanation

Use Fleming’s Left-Hand Rule:

 First finger (field): into the page

 Second finger (current/velocity): to the right

 Thumb (force): points upward

Since it’s a proton (positive charge), the force direction given by the rule is correct.

Final Answer: \(\boxed{\text{Upward}}\)

Example:

An electron is moving upward into a region where the magnetic field is from left to right. In which direction does the electron experience a force?

▶️ Answer/Explanation

Use Fleming’s Left-Hand Rule for a positive charge first:

First finger (field): left to right

Second finger (velocity): up

Thumb (force): out of the page (towards you)

But the particle is an electron (negative charge), so the force is in the opposite direction.

Final Answer: \(\boxed{\text{Into the page}}\)

 Lorentz Force

The Lorentz force is the total force experienced by a charged particle when it moves through both an electric field and a magnetic field.

Lorentz Force Equation

\(\vec{F} = q\vec{E} + q(\vec{v} \times \vec{B})\)

Where:

  • \(\vec{F}\) = total force on the particle (N)
  • \(q\) = charge of the particle (C)
  • \(\vec{E}\) = electric field (V/m)
  • \(\vec{v}\) = velocity of the particle (m/s)
  • \(\vec{B}\) = magnetic field (T)
  • \(\times\) = vector cross product

 Special Cases

  • If only electric field is present: \(\vec{F} = q\vec{E}\)
  • If only magnetic field is present: \(\vec{F} = q(\vec{v} \times \vec{B})\)

 Direction of Magnetic Lorentz Force

Use the Right-Hand Rule for the magnetic part:

  • Point fingers in the direction of \(\vec{v}\) (velocity)
  • Curl fingers in the direction of \(\vec{B}\) (magnetic field)
  • Thumb gives the direction of \(\vec{F}\) (for positive charge)

 For negative charges, the force is in the opposite direction.

 Units

  • Force \(\vec{F}\): newton (N)
  • Electric field \(\vec{E}\): volts per metre (V/m)
  • Magnetic field \(\vec{B}\): tesla (T)

Example:

An electron (\(q = -1.6 \times 10^{-19}\ \text{C}\)) is moving with velocity \(\vec{v} = 2.0 \times 10^6\ \text{m/s}\) perpendicular to a magnetic field of \(0.5\ \text{T}\). What is the magnitude of the magnetic force?

▶️ Answer/Explanation

Use: \(F = qvB\sin\theta\)

\(\theta = 90^\circ \Rightarrow \sin\theta = 1\)

\(F = (1.6 \times 10^{-19})(2.0 \times 10^6)(0.5)\)

\(F = 1.6 \times 10^{-13}\ \text{N}\)

Since it’s an electron, direction is opposite to right-hand rule prediction.

Final Answer: \(\boxed{1.6 \times 10^{-13}\ \text{N}}\)

Example:

A charged particle moves through perpendicular electric and magnetic fields such that it passes through undeviated. If \(E = 300\ \text{V/m}\) and \(B = 0.2\ \text{T}\), what is the velocity of the particle?

▶️ Answer/Explanation

For the particle to be undeviated: \(qE = qvB \Rightarrow E = vB\)

\(v = \frac{E}{B} = \frac{300}{0.2} = 1500\ \text{m/s}\)

Final Answer: \(\boxed{1500\ \text{m/s}}\)

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