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

Answer:

B = 5.59x10⁹ T

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:

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

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

I hope it helps you!