Question

Calculate the magnitude of the particle accelerator's magnetic field that causes an ionized helium atom (+2q e) with a momentum of 4.8X1016kgxm/s to orbit in the horizontal plane with a 1.0km radius. +2]q el = 2(1.6Xx10x10-19C) Note: q alpha A. 0.50T OB. 1.0T OC. 1.5T D. 2.0T OE. 2.5T

Answer 1

Answer:

The magnetic field of the particle is 1.5 T.

(C) is correct option.

Explanation:

Given that,

Momentum of particle $p=4.8\times10^{-16}\ kg m/s$

Radius = 1.0 km

Charge of the particle $q=2\times1.6\times10^{-19}\ C$

We need to calculate the magnetic field

Using relation of radius of path in magnetic field

$r=\dfrac{mv}{qB}$

Here mv = p

$B=\dfrac{p}{qr}$

Put the value into the formula

$B=\dfrac{4.8\times10^{-16}}{1000\times2\times1.6\times10^{-19}}$

$B=1.5\ T$

Hence, The magnetic field of the particle is 1.5 T.