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A coil carrying current is placed in the plane of a uniform magnetic field.
It carries current 0.2A , No. of turns is 25, magnetic field is 2T, radius 20cm. The area of crossection of wire is 5 x 10-6 m2 and the density of electrons is 1029 /m3. What is the average force on eah electron?

Radhika Batra , 11 Years ago
Grade 11
anser 1 Answers
Askiitians Tutor Team

Last Activity: 4 Months ago

To find the average force on each electron in a coil carrying current placed in a magnetic field, we can use the principles of electromagnetism. The force experienced by a charged particle in a magnetic field can be calculated using the formula:

Understanding the Force on Electrons

The force \( F \) on a charged particle moving in a magnetic field is given by the equation:

F = q(v × B)

Where:

  • F is the force on the charge (in Newtons)
  • q is the charge of the particle (in Coulombs)
  • v is the velocity of the particle (in meters per second)
  • B is the magnetic field strength (in Tesla)

Step-by-Step Calculation

1. **Determine the Charge of an Electron**: The charge of a single electron is approximately \( 1.6 \times 10^{-19} \) Coulombs.

2. **Calculate the Current Density**: The current density \( J \) can be calculated using the formula:

J = \frac{I}{A}

Where:

  • I is the current (0.2 A)
  • A is the cross-sectional area of the wire (5 x 10-6 m2)

Substituting the values:

J = \frac{0.2}{5 \times 10^{-6}} = 4 \times 10^{4} \, \text{A/m}^2

3. **Calculate the Drift Velocity**: The drift velocity \( v_d \) of the electrons can be calculated using the formula:

v_d = \frac{J}{n \cdot q}

Where:

  • n is the density of electrons (1029 /m3)

Substituting the values:

v_d = \frac{4 \times 10^{4}}{10^{29} \cdot 1.6 \times 10^{-19}} \approx 2.5 \times 10^{-4} \, \text{m/s}

4. **Calculate the Force on Each Electron**: Now we can find the force on each electron using the magnetic field strength \( B = 2 \, \text{T} \):

F = q \cdot v_d \cdot B

Substituting the values:

F = (1.6 \times 10^{-19}) \cdot (2.5 \times 10^{-4}) \cdot (2) \approx 8 \times 10^{-23} \, \text{N}

Final Thoughts

The average force on each electron in the coil, when subjected to the specified current and magnetic field, is approximately \( 8 \times 10^{-23} \) Newtons. This force is quite small, which is typical for individual electrons, but it plays a crucial role in the overall behavior of the current in the coil.

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