Question

1. an electron (m = 9.11 × 10^{-31} kg) enters a downward magnetic field of 5.00 × 10^{-1} t with a velocity of 6.50 × 10^{6} m/s west. calculate the radius of the circular path it will follow once it is travelling within the magnetic field. (3 marks)

Explanation:

Step1: Recall the formula for radius in magnetic field

The formula for the radius \( r \) of the circular path of a charged particle in a magnetic field is \( r=\frac{mv}{qB} \), where \( m \) is the mass of the particle, \( v \) is the velocity, \( q \) is the charge of the particle, and \( B \) is the magnetic field strength. For an electron, \( q = 1.6\times10^{-19}\ C \).

Step2: Identify the given values

Given: \( m = 9.11\times 10^{-31}\ kg \), \( v=6.50\times 10^{6}\ m/s \), \( B = 5.00\times 10^{-1}\ T \), \( q = 1.6\times 10^{-19}\ C \)

Step3: Substitute the values into the formula

Substitute the values into \( r=\frac{mv}{qB} \):

$$ r=\frac{(9.11\times 10^{-31}\ kg)\times(6.50\times 10^{6}\ m/s)}{(1.6\times 10^{-19}\ C)\times(5.00\times 10^{-1}\ T)} $$

Step4: Calculate the numerator and the denominator

First, calculate the numerator: \( (9.11\times 10^{-31})\times(6.50\times 10^{6})=9.11\times6.50\times 10^{-31 + 6}=59.215\times 10^{-25}=5.9215\times 10^{-24} \)

Then, calculate the denominator: \( (1.6\times 10^{-19})\times(5.00\times 10^{-1})=1.6\times5.00\times 10^{-19- 1}=8.0\times 10^{-20} \)

Step5: Divide the numerator by the denominator

$$ r=\frac{5.9215\times 10^{-24}}{8.0\times 10^{-20}}=\frac{5.9215}{8.0}\times 10^{-24 + 20}=0.7401875\times 10^{-4}=7.40\times 10^{-5}\ m $$

Answer:

The radius of the circular path is \( \boldsymbol{7.40\times 10^{-5}\ m} \) (or approximately \( 7.4\times 10^{-5}\ m \) depending on significant figures).