Question

ULINIL In a laboratory experiment, a beam of electrons is accelerated from rest through a 124-V potential difference. The beam then enters a uniform magnetic field and follows a circular path of radius r = 17.3 cm in the field region. (a) What is the angle between the magnetic field and the electrons' velocity? 90 (b) What is the magnitude of the magnetic field? IT Entra sumber Submit Answer Viewing Saved Work Revert to Last Response View Next Question

Answer 1

a) Since the beam follows a circular path, the angle between magnetic field and velocity of electron must be 90 degrees.

b)Velocity of a charge after being accelerated in a potential difference is given by: v=(2qV/m)^0.5, where v is velocity of the charge, q is magnitude of charge, V is potential difference and m is mass.

Here,q=magnitude of charge on electron=1.6*10^-19 C, V=124 volts, m=mass of electron=9.11*10^-31 kg.

So, velocity of electron beam=[2*1.6*10^-19*124/(9.11*10^-31)]^0.5=6599737.21 m/s.

Now, radius of a charged particle moving in uniform magnetic field=r=mv/(qB), where m is mass of the charged particle,v is its velocity,q is charge and B is magnetic field.

Here,r=17.3 cm=0.173 m

So,B=mv/(qr)=9.11*10^-31*6599737.21/(1.6*10^-19*0.173)=2.17*10^(-4) tesla.=0.000217 T.