Question

If a charged particle is thrown perpendicular to a
region of crossed electric field (E) and magnetic field
(B) goes undeviated, then the speed of particle is

Answer 1

Answer:

v can be any value as it is always 90 to e and b

Answer 2

The velocity of the charged particle is given by:

$\boxed{v_y\hat j+\frac{E}{B}\hat k}$

Explanation:

Given

The Electric Field and the Magnetic Field are perpendicular to each other

If the Electric Field is $\vec E=E\hat i$ and the magnetic field is $\vec B=B\hat j$

Let the velocity of  the charged particle be

$\vec v=v_x\hat i+v_y\hat j+v_z\hat k$

If the charge on the particle is q then

The force on the particle

= Force due to electric field + force due to magnetic field

$\implies \vec F=q\vec E+q(\vec v \cross\vec B)$

$\implies \vec F=q(E\hat i+ (v_x\hat i+v_y\hat j+v_z\hat k)\cross B\hat j)$

$\implies \vec F=q(E\hat i+ v_xB\hat k i-v_z\hat i)$

$\implies \vec F=q(E-v_zB)\hat i+ qv_xB\hat k$

Now if the charged particle passes through the region undeviated then

The force will be zero

i.e. $q(E-Bv_z)=0\implies v_z=\frac{E}{B}$

And $v_x=0$

$v_y$ can have any value

Therefore, the velocity is

$\vec v=v_y\hat j+\frac{E}{B}\hat k$

Hope this answer is helpful.

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