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We know that a charge moving in a magnetic field is subject to the force:
F = q · v · B

But we also know that:
F = m · a

Therefore, it must be:
m · a = q · v · B

And solving for a:
a = q · v · B / m

Recall that for a proton:
q = 1.6×10⁻¹⁹ C
m = 1.673×10⁻²⁷ kg

Now, you can find:
a = 1.6×10⁻¹⁹ · 7.0 · 1.7 / 1.673×10⁻²⁷
   = 1.14×10⁹ m/s²

Hence, the acceleration of the proton is 1.14×10⁹ m/s².
shubhamchouhanvrVT

The acceleration of a proton moving with a speed of 7.0 m/s at right angles to a magnetic field of 1.7 t will be   [tex]a=1.14\times 10^9 \dfrac{m}{s^2}[/tex]

What will be the acceleration of the proton?

When change moves in the amagnetic field then the force on the charge will be given as

[tex]F=q\times v\times B[/tex]

Here

q = charge

v= velocity of the charge

B= magnetic field

Now from newtons law, we have

[tex]F= m\times a[/tex]

So by equating the forces

[tex]m\times a=q\times v\times B[/tex]

[tex]a=\dfrac{q\times v\times B}{m}[/tex]  

Now for proton

[tex]q=1.6\times 10^{-19}\ C[/tex]

[tex]m=1.673\times 10^{-27}\ kg[/tex]

By putting the values in the formula

[tex]a=\dfrac{1.6\times10^{-19}\times7\times 1.7}{1.673\times 10^{-27}}[/tex]

[tex]a=1.14\times 10^9\ \frac{m}{s}[/tex]    

Thus the acceleration of a proton moving with a speed of 7.0 m/s at right angles to a magnetic field of 1.7 t will be   [tex]a=1.14\times 10^9 \dfrac{m}{s^2}[/tex]

To know more about  charge in the magnetic field follow

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