Question

please answer
A single charged ion has a mass of 1.13× 10^ -23 . It is accelerated through a potential difference of 500 V . It enters a magnetic field of 0.4 T , moving perpendicular to the field . The radius of its path in the field is ?

A) 2.1 cm
B ) 2.1 mm
C ) 1.17 m
D ) 2.0 m​

Answer 1

Explanation:

B ) 2.1 mm

hope it will help you

Answer 2

The correct option to the following question is (a), i.e., is 2.1 cm

Given: mass = 1.13 × 10⁻²³kg

          potential difference = 500V

          Magnetic field = 0.4T

To find: radius of the path

Solution: The electron moves in a circular path when it enters the magnetic field.

The kinetic energy of an electron moving in a magnetic field is equal to the potential energy that the electron possesses.

Therefore, 1/2 mv² = eV ( where m is the mass of the charged ion, v is the velocity, e is the magnitude of the charged particle, V is the potential difference )

The velocity of the charged ion = eBr/m, putting this in the above equation

Therefore, 1/2 m(eBr/m)² = eV

r = $\sqrt{2mV/B^2e}$

  = √(2×1.13× 10⁻²³×500)/(0.4)²×1.6×10⁻¹⁹

r = 2.1cm

Therefore, the radius of the path is 2.1cm