Question

A $200MeV$ proton enters a region of the magnetic field of intensity $5T$. The magnetic field points from south to north and the proton in moving along the vertical. The value of the force acting on the proton will be

• A. Zero
• B. $1.8\times {10}^{-10}N$
• C. $3.2\times {10}^{-18}N$
• D. $1.6\times {10}^{-6}N$

Answer 1

The correct option is D $1.6\times {10}^{-6}N$

We Know ,

$F=qvB$ and $sin\theta =1$ as motion and field are perpendicular

and $v=\sqrt{\frac{2K}{m}}$ K=Kinetic Energy

$F=q\sqrt{\frac{2K}{m}}B$

Putting all the values,

$F=1.6\times {10}^{-19}\times \sqrt{\frac{2\times 200\times {10}^6\times 1.6\times {10}^{-19}}{1.67\times {10}^{-27}}}\times 5$

$F=1.6\times {10}^{-19+8}\times 2\times 5=1.6\times {10}^{-10}N$

Hence option $D$ is correct.