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Question: A beam of well collimated cathode rays travelling with a speed of \(5 \times 10 ^ { 6 } \mathrm {~ms...

A beam of well collimated cathode rays travelling with a speed of 5×106 ms−15 \times 10 ^ { 6 } \mathrm {~ms} ^ { - 1 } enter a region of mutually perpendicular electric and magnetic fields and emerge undeviated from this region. If ∣B∣=0.02T| \boldsymbol { B } | = 0.02 T, the magnitude of the electric field is

A

105Vm−110 ^ { 5 } \mathrm { Vm } ^ { - 1 }

B

2.5×108Vm−12.5 \times 10 ^ { 8 } \mathrm { Vm } ^ { - 1 }

C

D

2×103Vm−12 \times 10 ^ { 3 } \mathrm { Vm } ^ { - 1 }

Answer

105Vm−110 ^ { 5 } \mathrm { Vm } ^ { - 1 }

Explanation

Solution

Using eE = evB ⇒E=vB=5×106×0.02=105Vm−1\Rightarrow E = v B = 5 \times 10 ^ { 6 } \times 0.02 = 10 ^ { 5 } \mathrm { Vm } ^ { - 1 }