Question

A superconducting Nb3Sn magnet can achieve a peak magnetic field of 13.5 T in a magnet designed for use in the Large Hadron Collider. Find the maximum energy that a singly charged particle (for example, a proton or electron) can have if that field is maintainted around a circular ring of circumference 39 km. (Note: In reality, particle energies are about 35% less, because the peak field is not maintained through the ring.)

Answer 1

Given Information:

Magnetic field = B = 13.5 T

Circumference of circular ring = c = 39 km

Required Information:

Maximum energy = E = ?

Answer:

Maximum energy = 4×10⁻⁶ Joules

Step-by-step explanation:

A charged particle experiences a magnetic force due to the magnetic field is given by

F = qvB

Where q is the magnitude of charge, v is the velocity and B is the magnetic field.

Since the charged particle is moving in a circular ring, the centripetal force acting on it is given by

F = mv²/r

Where m is the mass of charged particle and r is the radius of the circular ring.

qvB = mv²/r

qB = mv/r

but mv = p that is linear momentum of the charged particle

qB = p/r

p = qBr

The energy is given by

E = pc

Substitute p = qBr

E = qBrc

Where c is the speed of light c = 2.99×10⁸ m/s

First find the radius,

c = 2πr

r = c/2π

r = 39×10³/2π

r = 6207 m

So the energy of the charged particle is

E = qBrc

E = (1.60×10⁻¹⁹)(13.5)(6207)(2.99×10⁸)

E = 4×10⁻⁶ Joules