Question

What magnetic field strength is needed in each magnet to steer protons around the ring with a speed of 4.5 × 107 m/s? Assume that the field is uniform inside the magnet, zero outside.

Answer 1

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

Particle accelerators, such as the Large Hadron Collider, use magnetic fields to steer charged particles around a ring. Consider a proton ring with 36 identical bending magnets connected by straight segments. The protons move along a 3.5m long circular arc as they pass through each magnet.

What magnetic field strength is needed in each magnet to steer protons around the ring with a speed of 4.5×10⁷ m/s? Assume that the field is uniform inside the magnet, zero outside.

Given Information:

Radius = r = 36*3.5/2π = 20.05 m

Speed = v = 4.5×10⁷ m/s

Required Information:

Magnetic field = B = ?

Answer:

Magnetic field = 23.35 mT

Explanation:

The force acting on the protons due to magnetic field is given by

F = qvB

Since the protons are moving around a circular ring, the corresponding centripetal force is given by

F = mv²/r

Equating the both forces

qvB = mv²/r

qB = mv/r

B = mv/rq

Where m is the mass of proton (1.67×10⁻²⁷ kg), v is the speed of proton and q is the charge on proton (1.609×10⁻¹⁹ C)

Therefore, the magnetic field is

B = (1.67×10⁻²⁷*4.5×10⁷)/(20*1.609×10⁻¹⁹)

B = 0.02335 T

or

B = 23.35 mT

Answer 2

1.82 T

Step-by-step explanation:

Here is the complete question

Particle accelerators, such as the Large Hadron Collider, use magnetic fields to steer charged particles around a ring. Consider a proton ring with 36 identical bending magnets connected by straight segments. The protons move along a 3.5-m-long circular arc as they pass through each magnet.

Part A

What magnetic field strength is needed in each magnet to steer protons around the ring with a speed of 3.5 × 107 m/s? Assume that the field is uniform inside the magnet, zero outside.

Solution

The magnetic force on the proton equals the centripetal force on it.

So, mv²/r = Bev.

So, the magnetic field strength, B = mv/re

Since we have 36 straight circular arcs of length 3.5 m, the circumference of the circle that contains it is C = 36 × 3.5 m = 126 m. Since C = 2πr, the radius of the circle is r = C/2π = 126/2π = 20 m

So, B = mv/re where m = mass of proton = 1.67 × 10⁻²⁷ kg, v = speed of proton = 3.5 × 10⁷ m/s , e = proton charge = 1.609 × 10⁻¹⁹ C and r = 20 m

B = 1.67 × 10⁻²⁷ kg (3.5 × 10⁷ m/s)/(20 m × 1.609 × 10⁻¹⁹ C) = 0.182 × 10¹ = 1.82 T