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4.Moving Charges and Magnetism
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A beam of well collimated cathode rays travelling with a speed of $5 \times {10^6}\,m{s^{ - 1}}$ enter a region of mutually perpendicular electric and magnetic fields and emerge undeviated from this region. If $| B |=0.02\; T$, the magnitude of the electric field is
A
${10^5}\,V{m^{ - 1}}$
B
$2.5 \times {10^8}\,V{m^{ - 1}}$
C
$1.25 \times {10^{10}}\,V{m^{ - 1}}$
D
$2 \times {10^3}\,V{m^{ - 1}}$
Solution
(a) Using $eE = evB$
$ \Rightarrow E = vB = 5 \times {10^6} \times 0.02 = {10^5}\,V{m^{ – 1}}$
Standard 12
Physics
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