Question

A tightly wound long solenoid carries a current of 1.5 A. An electron is executing uniform circular motion inside the solenoid with a time period of 75 ns. The number of turns per meter in the solenoid is ________________.

Hint:

The magnetic field in a solenoid is directly proportional to the current and the number of turns per unit length. Use the magnetic force on the electron to relate the field and the motion.

Answer 1

Step 1: Understand the Given Information

The problem provides the following information:

• The current in the solenoid is 1.5 A.
  
• An electron is executing uniform circular motion inside the solenoid with a time period of 75 ns (75 × 10^-9 seconds).
  
• The number of turns per meter in the solenoid is to be determined.
  

Step 2: Formula for Magnetic Field in the Solenoid

The magnetic field inside a long solenoid is given by the formula:
B = μ₀ n I

where:
- B is the magnetic field strength,
- μ₀ is the permeability of free space (4π × 10^-7 T·m/A),
- n is the number of turns per meter,
- I is the current in the solenoid.
Substituting the given value of I = 1.5 A, we have:
B = μ₀ n × 1.5

Step 3: Force on the Electron in Circular Motion

The force on an electron moving in a magnetic field is given by the Lorentz force formula:
F = e v B

where:
- e is the charge of the electron (1.6 × 10^-19 C),
- v is the velocity of the electron,
- B is the magnetic field inside the solenoid.
For an electron in uniform circular motion, the magnetic force provides the centripetal force:
F = m v² / r where:
- m is the mass of the electron (9.11 × 10^-31 kg),
- v is the velocity of the electron,
- r is the radius of the circular motion.
Equating the two forces, we get:
e v B = m v² / r
Simplifying, we find the velocity of the electron:
v = e B r / m

Step 4: Relate Time Period to Magnetic Field and Radius

The time period T is the time taken for one complete revolution of the electron, which is related to the velocity and the radius by:
T = 2π r / v
Substituting the expression for v, we get:
T = 2π m / e B Rearranging for B, we have:
B = 2π m / e T Substituting the known values m = 9.11 × 10^-31 kg, e = 1.6 × 10^-19 C, and T = 75 × 10^-9 s, we get:
B = (2π × 9.11 × 10^-31) / (1.6 × 10^-19 × 75 × 10^-9) = 1.58 × 10^-3 T

Step 5: Calculate the Number of Turns per Meter

Now that we know the magnetic field B, we can substitute it into the equation for the magnetic field inside the solenoid to find n:
B = μ₀ n I
Substituting the known values B = 1.58 × 10^-3 T, μ₀ = 4π × 10^-7 T·m/A, and I = 1.5 A, we get:
1.58 × 10^-3 = (4π × 10^-7) n × 1.5
Solving for n, we find:
n = (1.58 × 10^-3) / ((4π × 10^-7) × 1.5) = 250 turns/m

Conclusion

The number of turns per meter in the solenoid is 250.

Answer 2

Step 1: Understand the magnetic field inside the solenoid.
The magnetic field inside a solenoid is given by the formula: \[ B = \mu_0 n I, \] where \( B \) is the magnetic field, \( \mu_0 \) is the permeability of free space, \( n \) is the number of turns per meter (turns density), and \( I \) is the current flowing through the solenoid.

Step 2: Determine the force on the electron.
The electron is moving in a circular path inside the solenoid, so the magnetic force provides the centripetal force. The magnetic force on an electron moving with velocity \( v \) in a magnetic field \( B \) is: \[ F_{\text{mag}} = e v B, \] where \( e \) is the charge of the electron and \( v \) is its velocity.
The centripetal force for uniform circular motion is: \[ F_{\text{centripetal}} = \frac{m v^2}{r}, \] where \( m \) is the mass of the electron and \( r \) is the radius of the circular path.

Equating the magnetic force and centripetal force: \[ e v B = \frac{m v^2}{r}. \] Simplifying: \[ B = \frac{m v}{e r}. \] Step 3: Relate the time period to velocity.
The time period \( T \) of the electron's circular motion is given by: \[ T = \frac{2\pi r}{v}. \] Thus, the velocity \( v \) is: \[ v = \frac{2\pi r}{T}. \] Substituting this into the equation for the magnetic field: \[ B = \frac{m \cdot \frac{2\pi r}{T}}{e r} = \frac{2\pi m}{e T}. \] Step 4: Solve for the number of turns per meter \( n \).
We now substitute the expression for \( B \) into the equation for the magnetic field in the solenoid: \[ \mu_0 n I = \frac{2\pi m}{e T}. \] Solving for \( n \): \[ n = \frac{2\pi m}{\mu_0 e T I}. \] Substitute the known values:
- \( m = 9.11 \times 10^{-31} \, \text{kg} \) (mass of the electron),
- \( e = 1.6 \times 10^{-19} \, \text{C} \) (charge of the electron),
- \( T = 75 \times 10^{-9} \, \text{s} \) (time period),
- \( I = 1.5 \, \text{A} \) (current),
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \).
Substituting these values gives: \[ n \approx 250 \, \text{turns per meter}. \] Final Answer:
\[ \boxed{250}. \]