Question

Problem 22.14 A6.70 - C particle moves through a region of space where an electric field of magnitude 1300 N/C points in the positive direction, and a magnetic field of magnitude 1.25 T points in the positive z direction.

Answer 1

Given:

q = 6.7 micro coulomb

E = 1300 N/C along +ve x-axis

B = 1.25 T along +ve z-axis

F = along +ve x-axis

6.25 * 10-3N

v = velocity = ?

Let the velocity v expressed in vector form as $\vec{v}=v_{x}\hat{i}+v_{y}\hat{j}$ .

The net force on the particle is

$\vec{F}=q(\vec{E}+(\vec{v}\times \vec{B}))$

$\Rightarrow (\vec{v}\times \vec{B}) = \frac{\vec{F}}{q}-\vec{E}$

6.25 * 10 = ((t + 3) x 1.25) = ( 6.7 * 10-6 - 1300)

$\Rightarrow -1.25v_{x}\hat{j}+1.25v_{y}\hat{i} = -367.16\hat{i}$

Therefore, comparing on both side,

we get the value of

$v_{x}=0$

and

$v_{y}=\frac{-367.16}{1.25}=-293.73m/s$.

Therefore, the velocity of the particle in x-y plane is (0, −293.73m/s). [answer]