An electron with velocity 3 * 10^7 ms-1 describes a circular path in a uniform megnatic field of 0.2T, perpendicular to it. Calculate the radius of path.
R=mv/eB
m =9.1•10 ⁻³¹kg
e =1.6•10⁻¹⁹ C.
To calculate the radius of the circular path of an electron moving in a uniform magnetic field, we can use the formula for the magnetic force acting on a charged particle moving perpendicular to the magnetic field.
The formula for the magnetic force (F) acting on a charged particle moving perpendicular to a magnetic field is given by:
F = qvB
Where:
F: Magnetic force acting on the electron
q: Charge of the electron (1.6 × 10^(-19) C)
v: Velocity of the electron (3 × 10^7 m/s)
B: Magnetic field strength (0.2 T)
In this case, the magnetic force provides the required centripetal force to keep the electron moving in a circular path. The centripetal force (Fc) is given by:
Fc = (mv^2) / r
Where:
m: Mass of the electron (9.1 × 10^(-31) kg)
v: Velocity of the electron (3 × 10^7 m/s)
r: Radius of the circular path
Since the centripetal force is equal to the magnetic force, we can equate the two equations:
(qvB) = (mv^2) / r
Now, we can solve this equation for the radius (r):
r = (mv) / (qB)
Plugging in the given values:
r = ((9.1 × 10^(-31) kg) × (3 × 10^7 m/s)) / ((1.6 × 10^(-19) C) × (0.2 T))
Calculating this expression:
r ≈ 0.028 m
Therefore, the radius of the electron's circular path is approximately 0.028 meters.