A proton and an alpha particle are moving perpendicular to a uniform magnetic field with the same speed.If the velocity of the proton is at an angle of 20° with respect to the magnetic field direction,at what angle to the magnetic field direction would the alpha particle be moving if it experiences the same magnitude acceleration as the proton? The alpha particle has 4 times the mass of the proton and twice its charge.
A)43°
B)10°
C)62°
D)9.8°

Question

Quizplus · Accepted Answer

The acceleration of a charged particle in a magnetic field is given by:$$a = qvB/m$$where:* q is the charge of the particle* v is the speed of the particle* B is the strength of the magnetic field* m is the mass of the particleSince the proton and alpha particle are moving with the same speed and experiencing the same magnitude acceleration, we can set their accelerations equal to each other:$$q_pv_pB/m_p = q_\alpha v_\alpha B/m_\alpha$$where:* q_p is the charge of the proton* v_p is the speed of the proton* m_p is the mass of the proton* q_\alpha is the charge of the alpha particle* v_\alpha is the speed of the alpha particle* m_\alpha is the mass of the alpha particleWe can simplify this equation by dividing both sides by B:$$q_pv_p/m_p = q_\alpha v_\alpha/m_\alpha$$Since the alpha particle has 4 times the mass of the proton and twice its charge, we can substitute these values into the equation:$$(1)(v_p)/(1) = (2)(v_\alpha)/(4)$$Solving for v_\alpha, we get:$$v_\alpha = 2v_p$$Now we can use the fact that the proton's velocity is at an angle of 20° with respect to the magnetic field direction to find the angle at which the alpha particle is moving. The angle between the velocity vector and the magnetic field vector is given by:$$	heta = 	an^{-1}(v_\perp/v_\|)$$where:* v_\perp is the component of the velocity vector perpendicular to the magnetic field* v_\| is the component of the velocity vector parallel to the magnetic fieldFor the proton, we have:$$	heta_p = 	an^{-1}(v_p\sin 20^\circ/v_p\cos 20^\circ) = 20^\circ$$For the alpha particle, we have:$$	heta_\alpha = 	an^{-1}(2v_p\sin 20^\circ/2v_p\cos 20^\circ) = 43^\circ$$Therefore, the correct answer is A.

A Proton and an Alpha Particle Are Moving Perpendicular to a Uniform