To determine the specific charge of cathode rays when they are subjected to crossed electric and magnetic fields, we need to use the condition where the beam experiences no deflection. This happens when the force due to the electric field and the force due to the magnetic field balance each other out.
Given:
• Electric field EE
• Magnetic field BB
• The velocity of the cathode rays is vv
• The specific charge of the cathode ray is em\dfrac{e}{m}, where ee is the charge and mm is the mass.
Forces Acting on the Cathode Ray:
1. Force due to Electric Field: The force FEF_E due to the electric field is given by:
FE=eEF_E = eE
where ee is the charge on the cathode ray and EE is the electric field strength.
2. Force due to Magnetic Field: The force FBF_B due to the magnetic field is given by the Lorentz force:
FB=evBF_B = evB
where vv is the velocity of the cathode ray and BB is the magnetic field strength.
Condition for No Deflection:
For the beam to remain undeflected, the forces due to the electric and magnetic fields must be equal and opposite:
FE=FBF_E = F_B eE=evBeE = evB
Simplifying this, we get:
E=vBE = vB
Solving for the Specific Charge:
From the equation E=vBE = vB, we can solve for the velocity vv of the cathode ray:
v=EBv = \dfrac{E}{B}
The force FEF_E (due to the electric field) can also be written in terms of the specific charge em\dfrac{e}{m}, as the force is related to the acceleration aa of the beam by FE=maF_E = ma. Since FE=eEF_E = eE and the acceleration aa is related to the velocity vv, we can write:
FE=em⋅m⋅aF_E = \dfrac{e}{m} \cdot m \cdot a
Now, substituting v=EBv = \dfrac{E}{B} into the equation, we find the expression for the specific charge em\dfrac{e}{m}:
em=B22vE2\dfrac{e}{m} = \dfrac{B^2}{2vE^2}
Thus, the correct expression for the specific charge is:
Answer: A. B22vE2\dfrac{B^2}{2vE^2}
