Question

When a charged particle moves at an angle of 27.2 degrees with respect to a magnetic field, it experiences a magnetic force of magnitude F. At what angle (less than 90 degrees) with respect to this field will this particle, moving at the same speed, experience a magnetic force of magnitude 2F?

Answer 1

force due to magnetic field =

here q = charge of the particle

= velocity of the particle

= magnetic field

F = qvB sin ()

so F₂/F₁ = sin (₂)/sin (₁)

here ₁ = 27.2⁰

₂ = ?

F₂/F₁ = 2

solving

sin ₂ = 66.1⁰

Answer 2

Calculation:

The magnetic force on a charged particle moving at an angle θ to a magnetic field is given by:

$$F=qvB\sin \theta$$

We are given:

1. At θ₁ = 27.2°, the force is $F$.

2. We need to find θ₂ where the force is $2F$, with the same $q$, $v$, and $B$.

Set up the ratio of the forces:

$$\frac{2F}{F}=\frac{qvB\sin {\theta }_2}{qvB\sin {\theta }_1}$$

Simplify:

$$2=\frac{\sin {\theta }_2}{\sin (27.2\mathrm{°})}$$

Solve for $\sin {\theta }_2$:

$$\sin {\theta }_2=2\cdot \sin (27.2\mathrm{°})$$$$\sin {\theta }_2=2\cdot 0.457$$$$\sin {\theta }_2=0.914$$

Find θ₂:

$${\theta }_2={\sin }^{-1}(0.914)$$$${\theta }_2\approx 61.6\mathrm{°}$$

Thus, the angle is 61.6 degrees