Question

A proton (or charged particular) moving with velocity v is acted upon by electric field E and magnetic field B. The proton will move undeflected if
A. \[E\]is perpendicular to \[B\].
B. \[E\]is parallel to \[v\] and perpendicular to \[B\].
C. \[E,B,v\] are mutually perpendicular to \[B\] and \[v = \dfrac{E}{B}\]
D. \[E\] and \[B\] both are parallel to \[v\].

Answer 1

Hint: In the given question, we need to determine the condition for which the proton will move undeflected. For this, we need to use the concept that the magnitude of force due to electric field is equal to the magnitude of force due to magnetic field but these two forces are opposite in direction to get the desired result.





Complete answer:
We know that the magnitude of force due to the electric field is equal to the magnitude of force due to the magnetic field but these two forces are opposite in direction.
Mathematically, it is given by
\[\left| {{{\vec F}_e}} \right| = \left| {{{\vec F}_m}} \right|\]
Here, \[{F_e}\] is the force due to an electric field and \[{F_m}\] is the force due to a magnetic field.
Thus, we get
\[qE = q(vB)\]
Here, \[q\] is the charge, \[v\] is the velocity, \[E\] is an electric field and \[B\] is a magnetic field of intensity.
Thus, we get
$qE = q {v \times B}$
$E = {v \times B}$
Hence, the proton will move undeflected if \[E,B,v\] are mutually perpendicular to \[B\] and \[v = \dfrac{E}{B}\].
Therefore, the correct option is (C).


Note:Many students make mistakes in analyzing the concept of the magnitude of force due to electric field is equal to the magnitude of force due to magnetic field but these two forces are opposite in direction. So, it is essential to get the correct result.