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Question: \- In an expt, a beam of electron passes undeviated through mutually perpendicular elec. and mag fie...

- In an expt, a beam of electron passes undeviated through mutually perpendicular elec. and mag fields of respective strength E=7.2×106  NC1E = 7.2 \times {10^6}\;{\rm{N}}{{\rm{C}}^{{\rm{ - 1}}}} and . The velocity of the electron is
A. 17.3×107  ms117.3 \times {10^7}\;{\rm{m}}{{\rm{s}}^{{\rm{ - 1}}}}
B 3×106  ms13 \times {10^6}\;{\rm{m}}{{\rm{s}}^{{\rm{ - 1}}}}
C 2×106  ms12 \times {10^6}\;{\rm{m}}{{\rm{s}}^{{\rm{ - 1}}}}
D. 6×106  ms16 \times {10^6}\;{\rm{m}}{{\rm{s}}^{{\rm{ - 1}}}}

Explanation

Solution

This question is based on the electric and magnetic field. First, use the formula of electric force and magnetic force. The electric force is dependent on charge and strength of electric field and magnetic force is dependent on charge, velocity and strength of magnetic field. Now, equate both electric and magnetic force formulas to get the desired result and get the velocity of the electron.

Complete step by step answer:
Given: The strength of electric field is E=7.2×106  NC1E = 7.2 \times {10^6}\;{\rm{N}}{{\rm{C}}^{{\rm{ - 1}}}} and the strength of magnetic field is B=2.4  TB = 2.4\;{\rm{T}}.
First we have to find the force generated due to electric field
We know that the electric force is directly related to the electric field and charge. The electric force is changed only due to the strength of the electric field because the charge on the electron remains constant.
So, by using the relation between force and charge we get,
FE=qE{F_E} = qE
Here, qq is the charge of an electron and EE is the electric field.
So we can write as,
FE=eE........(i){F_E} = eE........\left( {\rm{i}} \right)
Now, we can write the relation between the magnetic field and force we get,
FB=qvB     FB=veB......(ii) {F_B} = qvB\\\ \implies {F_B} = veB......\left( {{\rm{ii}}} \right)
Here, BB is the magnetic field, and vv is the velocity of the electron.
Now, equate the two equations and write in the form of velocity we get,
eE=evB     v=EB eE = evB\\\ \implies v = \dfrac{E}{B}
Substitute the values in above equation we get,
v=7.2×106  NC12.4  T     v=3×106  ms1 v = \dfrac{{7.2 \times {{10}^6}\;{\rm{N}}{{\rm{C}}^{{\rm{ - 1}}}}}}{{2.4\;{\rm{T}}}}\\\ \implies v = 3 \times {10^6}\;{\rm{m}}{{\rm{s}}^{{\rm{ - 1}}}}
Therefore, the velocity of the electron is 3×106  ms13 \times {10^6}\;{\rm{m}}{{\rm{s}}^{{\rm{ - 1}}}}.

Thus, the correct option is (B).

Note:
In this question, we have knowledge of electric current and generation of electric current. If we increase more and more current to remove the resistance, which is practically not possible because if we increase more current, then there is a chance of an electric device to burst. There is some limit that an electric device can bear that amount of current.