Question

"An alpha particle (He2+) with a velocity of 2.6x106m/scrosses a uniform magnetic field at an angle of 37.0°to the field lines and experiences a magnetic force of 1.4x10-3N. Find the strength of the magnetic field."

Answer 1

B = 5.59x10⁹ T

Step-by-step explanation:

The magnetic force (F), on a the alpha particle with charge (q) that is moving at velocity (v) as the cross product of the velocity and magnetic field (B) is:

`F = qvBsin(\theta)`

We have:

F = 1.4x10⁻³ N

v = 2.6x10⁶ m/s

θ = 37.0°

q = 2*p = 2*1.6x10⁻¹⁹ C

Hence, the strength of the magnetic field is:

`B = (F)/(qvsin(\theta)) = (1.4 \cdot 10^(-3))/(1.6 \cdot 10^(-19) C*2.6 \cdot 10^(6)*sin(37)) = 5.59 \cdot 10^(9) T`

Therefore, the strength of the magnetic field is 5.59x10⁹ T.

I hope it helps you!