A beam of cathode rays remains undeflected when crossing mutually perpendicular electric and magnetic fields of magnitudes E and B. If the rays were accelerated through potential difference V, what is their specific charge e/m?
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Question

A beam of cathode rays remains undeflected when crossing mutually perpendicular electric and magnetic fields of magnitudes E and B. If the rays were accelerated through potential difference V, what is their specific charge e/m?

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G1 Physics · Accepted Answer

Step 1: In crossed fields with no deflection, eE = evB, so particle speed v = E/B. Step 2: From accelerating voltage V, eV = (1/2)mv². Step 3: Divide by mass and substitute v = E/B: e/m = v²/(2V) = E²/(2VB²). Final answer: (D) E²/(2VB²).