Protons ( q p = 1.60 × 10 -19 C, m p = 1.67 × 10 -27 kg) areaccelerated from res

Question

Protons (qp = 1.60 × 10-19C, mp = 1.67 × 10-27 kg) areaccelerated from rest through a potential differenceVa and emerge moving in the +ydirection with a speed v0 = 3 ×106 m/s. After passing through a field-free region, theyenter at the origin a region (y > 0) with a uniformmagnetic field B = 2.5 T in the +z direction (outof the page). They then execute a semicircular trajectory andemerge from the magnetic field region at distancex1 from the origin.

(a) What is the magnitude of the potential differenceVa?

|Va| = V    

(b) Calculate the value of x1 at which theprotons exit the magnetic field region.

x1 = m    

(c) Suppose instead deuterons (qD =qp, mD = 2mp)are accelerated from rest through the same potentialVa and enter the same magnetic field. Calculatethe position x1' at which deuterons exit themagnetic field region.

x1' = m

Explanation / Answer

Given that qp = 1.60 × 10-19 C mp = 1.67 × 10-27 kg v0 = 3 × 106 m/s. B = 2.5 T (a) What is the magnitude of the potential differenceVa From conservation mechanical energy we can write           K + U = 0 we get          0.5mvo2 - qVa = 0 ==> Va =  mvo2 / 2 q                = 1.67 x 10-27 * (3 x 106)2 / 2 * 1.6 x 10-19                = 4.696 x 104 V (b) Calculate the value of x1 at whichthe protons exit the magnetic field region.        We know that radius of thepath is               r = m v / q B                  = [ m / qB ] * [ (2qVa/m)]                  = [ 1 / B ] * [ 2 mVa / q ] Thus x1 = 2r = [ 2 / B ] * [ (2mVa / q) ]                     = [ 2 / 2.5] * [ ( 2 * 1.67 x 10-27 * 4.696x 104 / 1.6 x 10-19 ) ]                     = 0.8 *   [ 3.130 x 10-2 ]                     = 0.02504 m (c) Suppose instead deuterons (qD =qp, mD = 2mp)are accelerated from rest through the same potentialVa and enter the same magnetic field. Calculatethe position x1' at which deuterons exit themagnetic field region.          x1 = 2r= [ 2 / B ] * [ (2 mDVa /qD) ]                     = [ 2 / 2.5] * [ ( 2 * 2*1.67 x 10-27 *4.696 x 104 / 1.6 x 10-19 ) ]                     = 0.8 *   [ 4.427 x 10-2 ]                     = 0.03542 m

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.