Question

Now Safiya and Amy try a homework problem. Consider an electron moving with a speed of 8.70 106 m/s at right angles to a constant magnetic field, with B = 1.7 T. What is the radius of the circular path of the electron? r = m

Answer 1

Answer:

Radius of circular path will be $29.10\times 10^{-6}m$

Explanation:

Mass of electron $m=9.1\times 10^{-31}kg$

Charge on electron $q=1.6\times 10^{-19}C$

Speed of electron is given $v=8.7\times 10^6m/sec$

Magnetic field is given B = 1.7 T

We have to find the radius of the circular path

Radius of circular path is equal to $r=\frac{mv}{qB}$ , here m is mass v is velocity q is charge and B is magnetic field

So $r=\frac{9.1\times 10^{-31}\times 8.7\times 10^6}{1.6\times 10^{-19}\times 1.7}=29.10\times 10^{-6}m$

So radius of circular path will be $29.10\times 10^{-6}m$