Question

What is the strength of a magnetic field if a proton moving at 350 m/s experienced a magnetic force of 8 N while moving along the field?

Answer 1

Explanation:

$f= qvb \sin( \beta )$

$f = 8n \\ v = 350ms {}^{ - 1} \\ \beta = \: 90 {}^{0 } \\ b = \frac{f}{qv} \\ \\ b = \frac{8}{350 \times 1.6 \times 10 {}^{ - 19} }$

B = 1.42857 × 10¹⁹

Answer 2

Answer:

proton is moving along the field so the angle between velocity and the magnetic field is zero  so the particle experience no force

Explanation:

• When a particle with charge $q$ moving with velocity $V$ through the magnetic field by making an angle $\alpha$ with the magnetic field, the particle experiences a force called Lorentz force, which is given by the formula

                                             $F=qVB sin\alpha$

From the question, we have given

velocity of proton $V=350m/s$

the magnetic force experienced by proton $F=8N$

the charge of the proton $q=1.6*10^{-19}C$

proton is moving along the field so the angle between velocity and the magnetic field is $\alpha =0$

at this angle, force should be zero since $sin0=0$

so the question is wrong.