Question

A proton traveling with speed 2 × 105 m/s in the -y direction passes through a region in which there is a uniform magnetic field of magnitude 0.6 T in the -z direction. You want to keep the proton traveling in a straight line at constant speed. To do this, you can turn on an apparatus that can create a uniform electric field throughout the region. What electric field should you apply?

Answer 1

Answer:

$\vec{E} =  1.2\times 10^5(-i)$

Explanation:

Given that

Speed ,v= 2 x 10⁵ m/s ( - y direction)

B= 0.6 T (- z direction)

The resultant force on the proton given as

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

F= m a

For uniform motion acceleration should be zero.

F = 0

$0=q.\vec{E}+ q.(\vec{v}\times \vec{B})$

$0=\vec{E}+ (\vec{v}\times \vec{B})$

$0=\vec{E}+2\times 10^5(-j) \times 0.6(-k)$

$\vec{E} =- 2\times 10^5(-j) \times 0.6(-k)$

$\vec{E} =-1.2\times 10^5(i)$

$\vec{E} = 1.2\times 10^5(-i)$

Electric filed should be apply in the negative x direction.