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A solenoid has 2×10^4 turns per meter and has diameter 10 cm .An electron beam having K.E 100 Kev passes without touching walls of solenoid then find current in the solenoid. Electron beam make 30 degree with axis of the solenoid.?
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Solution:

Given:
Number of turns per meter (N) = 2×10^4
Diameter of the solenoid (d) = 10 cm = 0.1 m
Kinetic Energy of the electron beam (KE) = 100 keV = 100 × 10^3 eV

To find:
Current in the solenoid.

Step 1: Calculating the magnetic field inside the solenoid

The magnetic field inside a solenoid is given by the formula:

B = μ₀ × N × I

Where:
B is the magnetic field
μ₀ is the permeability of free space (4π × 10^-7 Tm/A)
N is the number of turns per meter
I is the current in the solenoid

We need to find the current (I), so we rearrange the formula as:

I = B / (μ₀ × N)

Step 2: Calculating the magnetic field

The magnetic field at a point due to a moving charged particle is given by the formula:

B = (μ₀ / 4π) × (q × v × sinθ) / r²

Where:
B is the magnetic field
μ₀ is the permeability of free space (4π × 10^-7 Tm/A)
q is the charge of the particle (in this case, the charge of an electron = -1.6 × 10^-19 C)
v is the velocity of the particle (in this case, the velocity of the electron beam)
θ is the angle between the velocity vector and the magnetic field direction (30° in this case)
r is the distance between the electron beam and the point where we want to calculate the magnetic field (in this case, the radius of the solenoid)

Step 3: Calculating the velocity of the electron beam

The kinetic energy of the electron beam can be written as:

KE = (1/2) × m × v²

Where:
KE is the kinetic energy
m is the mass of the electron (9.1 × 10^-31 kg)
v is the velocity of the electron beam

Rearranging the formula, we find:

v = √(2 × KE / m)

Step 4: Calculating the magnetic field inside the solenoid

Using the formula for the magnetic field due to a moving charged particle, we can calculate the magnetic field inside the solenoid. Since the electron beam is passing through the solenoid without touching its walls, we can assume that the distance (r) is equal to the radius of the solenoid (0.05 m). Plugging in the values:

B = (4π × 10^-7 Tm/A) × (-1.6 × 10^-19 C) × (√(2 × 100 × 10^3 eV / (9.1 × 10^-31 kg))) × sin(30°) / (0.05 m)²

Simplifying the expression:

B = (4π × 10^-7 Tm/A) × (-1.6 × 10^-19 C) × (√(2 × 100 × 10^3 eV / (9.1 × 10^-31 kg))) ×
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A solenoid has 2×10^4 turns per meter and has diameter 10 cm .An electron beam having K.E 100 Kev passes without touching walls of solenoid then find current in the solenoid. Electron beam make 30 degree with axis of the solenoid.?
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