Question

13. A proton moves at 7.50×107 m/s perpendicular to a magnetic field. The field causes the proton to travel in a circular path of radius 0.800 m. What is the field strength

Answer 1

The magnetic force acting on a charged particle moving perpendicular to the field is:

$F_{b}$ = qvB

$F_{b}$ is the magnetic force, q is the particle charge, v is the particle velocity, and B is the magnetic field strength.

The centripetal force acting on a particle moving in a circular path is:

$F_{c}$ = mv²/r

$F_{c}$ is the centripetal force, m is the mass, v is the particle velocity, and r is the radius of the circular path.

If the magnetic force is acting as the centripetal force, set $F_{b}$ equal to $F_{c}$ and solve for B:

qvB = mv²/r

B = mv/(qr)

Given values:

m = 1.67×10⁻²⁷kg (proton mass)

v = 7.50×10⁷m/s

q = 1.60×10⁻¹⁹C (proton charge)

r = 0.800m

Plug these values in and solve for B:

B = (1.67×10⁻²⁷)(7.50×10⁷)/(1.60×10⁻¹⁹×0.800)

B = 0.979T

Answer 2

The magnetic field strength of the proton is 0.979Tesla

The magnetic force acting on a charged particle moving perpendicular to the field is expressed using the equation.

Fm = qvB

The centripetal force traveled by the proton in a circular path is expressed as:

Fp = mv²/r

To get the field strength, we will equate both the magnetic force and the centripetal force as shown:

Fm = Fp

qvB = mv²/r

qB = mv/r

m is the mass of a proton

v is the velocity = 7.50×10⁷ m/s

m is the mass on the proton = 1.67 × 10⁻²⁷kg

q is the charge on the proton =  1.60×10⁻¹⁹C

r is the radius = 0.800m

Substitute the given parameters into the formula as shown:

$B=\frac{mv}{qr}\\B = \frac{1.67 \times 10^{-27} \times 7.5 \times 10^7}{1.60 \times 10^{-19} \times 0.8}$

$B=0.979Tesla$

On solving, the magnetic field strength of the proton is 0.979Tesla

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