An electron is revolving in a circular path of radius 2× 10^-10m with a speed 3×10^6m/s . The magnetic field at the centre of circular path will be (ans is 1.2T)?

Question

Answer

Calculating the Magnetic Field at the Centre of Circular Path of an Electron

Given Information:

Radius of circular path (r) = 2 x 10^-10m

Speed of electron (v) = 3 x 10^6m/s

Formula Used:

Magnetic field (B) = (μ0 * q * v) / (2 * r)

Calculation:

Electric charge of an electron (q) = -1.6 x 10^-19 C

Permeability of free space (μ0) = 4π x 10^-7 Tm/A

Substituting the given values in the formula:

B = (4π x 10^-7 Tm/A) * (-1.6 x 10^-19 C) * (3 x 10^6 m/s) / (2 * 2 x 10^-10m)

B = -3.84 x 10^-26 T

The magnitude of magnetic field at the centre of circular path is:

|B| = 3.84 x 10^-26 T

As the magnetic field is a vector quantity, it has both magnitude and direction. The direction of magnetic field can be determined using the right-hand rule.

Final Answer:

The magnetic field at the centre of circular path of an electron is 1.2T (direction not specified).

Note: The given answer of 1.2T is incorrect. The correct answer is -3.84 x 10^-26 T.

Answer

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