Question

A proton travels with a speed of 8. 34×10^5 m/s at an angle of 32. 3° with the direction of a magnetic field. the field exerts a force of 3. 00×10^−17 n on the proton. What is the strength of the field?

Answer 1

Magnetic field strength of the field is  4.2 × $10^{-4}$ T

What is Magnetic Field ?

The region of space where magnetic force is experienced is known as magnetic field.

The force on a particle of charge q moving at speed v at an angle Ф to a uniform magnetic field of flux density B is given as

F = BqvsinФ

Given that a proton travels with a speed of 8. 34×10^5 m/s at an angle of 32. 3° with the direction of a magnetic field. the field exerts a force of 3. 00×10^−17 N on the proton. What is the strength of the field?

The parameters are;

• Force F = 3.0 × 10^−17 N

• Speed V = 8.34 × 10^5 m/s

• Angle Ф =  32. 3°

• Magnetic field strength B = ?

• Proton charge = 1.6 x $10^{-19}$ C

Substitute all the parameters into the formula

F = BqvsinФ

3.0 × $10^{-17}$ = ( 1.6 x $10^{-19}$  x 8.34 x $10^{5}$ x sin 32.3 ) B

3.0 × $10^{-17}$  = 7.13 × $10^{-14}$ B

Make B the subject of formula

B = 3.0 × $10^{-17}$ / 7.13 × $10^{-14}$

B = 4.2 × $10^{-4}$ T

Therefore, Magnetic field strength of the field is  4.2 × $10^{-4}$ T

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