Question

A particle of charge equal to 10 times the charge of electrons revolves in a circle with frequency equal to 10 revolutions per second. Find the magnetic field at the centre of the circular path.

Hint:

The magnetic field in a circular path depends on the current, charge, and radius. For charged particles moving in a magnetic field, use the relationship between charge, frequency, and current.

Answer 1

The magnetic field at the centre of the circular path of a charged particle moving in a magnetic field is given by: \[ B = \frac{\mu_0 I}{2r} \] where \( \mu_0 \) is the permeability of free space, \( I \) is the current, and \( r \) is the radius of the circular path. The current \( I \) is related to the charge \( q \) and the frequency \( f \) of the particle as: \[ I = qf \] Given that the charge on the particle is 10 times the charge of an electron, \( q = 10e \), and the frequency is \( f = 10 \) revolutions per second, the current is: \[ I = 10e \cdot 10 = 100e \] The radius \( r \) of the circular path is related to the momentum \( p \) of the particle and the magnetic field \( B \) by the relation: \[ r = \frac{mv}{qB} \] Substitute \( I = qf \) into the magnetic field equation and solve for \( B \), noting that the specific numerical values for the electron's charge and mass will provide the final result.