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Question

A helium nucleus makes a full rotation in a circle of radius 0.8 meter in two second. The value of B (magnetic field) at the center of the circle will be


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Solution

Step 1. Given data

Helium nucleus makes a full rotation in a circle of radius 0.8

Time taken for one full rotation of the Helium nucleus is 2second

Step 2. Formula used

  1. I=2π
  2. B=μ0I2r
  3. f=ω2π
  4. f=1t

Where B is the magnetic field, μ0 is the permeability of the free space, I is the current, r is the radius, q is the charge of Helium atom, ω is the angular frequency of nth orbit and t is the time.

Step 3. Find the angular frequency

t=2 second

From equation (iii) and (iv) we get

1t=ω2πω=2πtω=π

Step 4. Find the current due to revolution of the electron

I is the current due to revolution of the electron.

r=0.8m

q=2×1.6×10-19C is the charge of Helium atom.

Then ,

I=2π=2×1.6×10-192=1.6×10-19A

Step 5. Find the value of magnetic field

B=μ0I2r=μ0×1.6×10-192×0.8=μ010-19=4π×10-7×10-19=1.26×10-25tesla

Hence the value of the magnetic field in the center of the circle is 1.26×10-25tesla.


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