Question

An electron at point A has a speed v0 of 1.41x106 m/s. Find (a) the magnitude and direction of the magnetic field that will cause the electron to follow a semi circular path from A to B, and

Answer 1

Answer:

(a). The magnitude of the magnetic field is $16.0\times10^{-5}\ T$

(b). The time required for the electron to move from A to B is $1.11\times10^{-7}\ sec$.

(c). The magnetic field is 0.2943 T.

Explanation:

Given that,

Speed $v= 1.41\times10^{6}\ m/s$

We need to calculate the magnetic field

Using formula of magnetic force and centripetal force

$F_{b}=F_{c}$

$qvB=\dfrac{mv^2}{r}$

$B=\dfrac{mv}{qr}$

Where, m = mass of electron

v = speed of electron

q = charge of electron

r = radius of circular path

Put the value into the formula

$B=\dfrac{9.1\times10^{-31}\times1.41\times10^{6}}{1.6\times10^{-19}\times5\times10^{-2}}$

$B=0.000160\ T$

$B=16.0\times10^{-5}\ T$

The direction of the magnetic field must be into the page by the right hand and curls finger rule.

(b).  The time required for the electron to move from A to B will be given as :

We need to calculate the time

Using formula of time

$t=\dfrac{d}{v}$

Where, d =  total distance traveled by an electron

v = speed of electron

Put the value into the formula

$t=\dfrac{\pi \times5\times10^{-2}}{1.41\times10^{6}}$

$t=1.11\times10^{-7}\ sec$

The time required for the electron to move from A to B is $1.11\times10^{-7}\ sec$.

(c). If the particle were a proton instead of an electron

We need to calculate the magnetic field

Using formula of magnetic field

$B=\dfrac{mv}{qr}$

Put the value into the formula

$B=\dfrac{1.67\times10^{-27}\times1.41\times10^{6}}{1.6\times10^{-19}\times5\times10^{-2}}$

$B=0.2943\ T$

The magnetic field is 0.2943 T.

Hence, (a). The magnitude of the magnetic field is $16.0\times10^{-5}\ T$

(b). The time required for the electron to move from A to B is $1.11\times10^{-7}\ sec$.

(c). The magnetic field is 0.2943 T.