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Chapter 19: Problem 29

The conversion between atomic mass units and kilograms is $$1 \mathrm{u}=1.66 \times 10^{-27} \mathrm{kg}$$ After being accelerated through a potential difference of \(5.0 \mathrm{kV},\) a singly charged carbon ion \(\left(^{12} \mathrm{C}^{+}\right)\) moves in a circle of radius \(21 \mathrm{cm}\) in the magnetic ficld of a mass spectrometer. What is the magnitude of the field?

Short Answer

Expert verified
Question: Calculate the magnitude of the magnetic field that causes a singly charged carbon ion to move in a circular path after being accelerated through a potential difference of 5.0 kV. The radius of the circular path is 21 cm. Answer: The magnitude of the magnetic field is 1.09 T.

Step by step solution

01

Identify given information

We are given: 1. Conversion between atomic mass units (u) and kilograms: 1 u = 1.66 * 10^(-27) kg 2. Potential difference (V) through which the ion is accelerated: 5.0 kV = 5.0 * 10^3 V 3. Mass of a singly charged carbon ion (m): 12 u = 12 * 1.66 * 10^(-27) kg 4. Radius of the circular path (r): 21 cm = 0.21 m 5. Charge of a singly charged carbon ion (q): +1 e = +1 * 1.60 * 10^(-19) C
02

Calculate the ion's final kinetic energy

Final kinetic energy: KE = qV We use the equation KE = qV, where KE is kinetic energy, q is the ion's charge, and V is the potential difference, to find the final kinetic energy of the ion. KE = (1 * 1.60 * 10^(-19) C)(5.0 * 10^3 V) = 8 * 10^(-16) J
03

Calculate the ion's final velocity

Relationship between kinetic energy and velocity: KE = (1/2)mv^2 We can rearrange the formula for kinetic energy, KE = (1/2)mv^2, to solve for the velocity (v): v = sqrt(2 * KE / m) where m is mass and KE is the kinetic energy we calculated in step 2. v = sqrt(2 * 8 * 10^(-16) J / (12 * 1.66 * 10^(-27) kg)) = 6.53 * 10^5 m/s
04

Calculate the magnetic field magnitude

Force acting on a charged particle in a magnetic field: F = qvB We can use the formula F = qvB, where F is the force acting on the ion, q is the ion's charge, v is the velocity, and B is the magnetic field magnitude. Since the ion is moving in a circular path due to the magnetic field, the centripetal force and magnetic force acting on it are equivalent: F_centr = F_magnetic => (mv^2) / r = qvB We can rearrange the formula to solve for the magnetic field magnitude (B): B = (mv^2) / (qr) where m is mass, v is the final velocity calculated in step 3, q is charge, and r is the radius of the circular path. B = ((12 * 1.66 * 10^(-27) kg)(6.53 * 10^5 m/s)^2) / ((1 * 1.60 * 10^(-19) C)(0.21 m)) = 1.09 T
05

Write the final answer

The magnitude of the magnetic field is 1.09 T.

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Most popular questions from this chapter

Two parallel wires in a horizontal plane carry currents \(I_{1}\) and \(I_{2}\) to the right. The wires each have length \(L\) and are separated by a distance \(d\). (a) What are the magnitude and direction of the ficld due to wire 1 at the location of wire \(2 ?\) (b) What are the magnitude and direction of the magnetic force on wire 2 due to this field? (c) What are the magnitude and direction of the field due to wire 2 at the location of wire \(1 ?\) (d) What are the magnitude and direction of the magnetic force on wire 1 due to this field? (e) Do parallel currents in the same direction attract or repel? (f) What about parallel currents in opposite directions?
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