Assume the region to the right of a certain plane contains a uniform magnetic fi

Question

Assume the region to the right of a certain plane contains a uniform magnetic field of magnitude

b = 1.03 mT

and the field is zero in the region to the left of the plane as shown in the figure below. An electron, originally traveling perpendicular to the boundary plane, passes into the region of the field.

2· -/14.28 points My Notes Ask Your Teacher Assume the region to the right of a certain plane contains a uniform magnetic field of magnitude b 1.03 mT and the field is zero in the region to the left of the plane as shown in the figure below. An electron, originally traveling perpendicular to the boundary plane, passes into the region of the field XxXXX (a) Determine the time interval required for the electron to leave the "field-filled" region, noting that the electron's path is a semicircle (b) Assuming the maximum depth of penetration into the field is 2.47 cm, find the kinetic energy of the electron ev Need Help?Read It

Explanation / Answer

Using Force balance on electron inside magnetic field is

Fc = Fm

m*V^2/r = q*V*B

r = mV/qB

V = q*B*r/m

Time period of motion will be

T = 2*pi/w

w = V/r

T = 2*pi*r/V

T = 2*pi*(mV/qB)/V

T = 2*pi*m/qB

Since we need the time required to leave the field, which will be half of time period, So

t = T/2 = pi*m/(qB)

Now using given values:

t = pi*9.1*10^-31/(1.6*10^-19*1.03*10^-3)

t = 17.35*10^-9 sec

Part B

Kinetic energy is given by:

KE = 0.5*m*V^2 = 0.5*m*(qBr/m)^2

KE = (qBr)^2/2m

Using given values:

KE = (1.6*10^-19*1.03*10^-3*2.47*10^-2)^2/(2*9.1*10^-31)

KE = 9.10*10^-18 J = (9.10*10^-18)/(1.6*10^-19) = 56.875 eV

Please Upvote.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---