Question

find the linear momentum of a proton moving on a circular path of radius 0.1m in a magnetic field of 0.05 weber/m2

Answer 1

Momentum of proton will be $8\times 10^{-22}kgm/sec$

Explanation:

We have given radius of the circular path r = 0.1 m

Magnetic filed B = $0.05weber/m^2$

Charge on proton will be $q=1.6\times 10^{-19}C$

Mass of proton $m=1.67\times 10^{-27}kg$

On circular path centripetal force will be equal to magnetic force on the proton

So $qvB=\frac{mv^2}{r}$

$v=\frac{Bqr}{m}=\frac{0.05\times 1.6\times 10^{-19}\times 0.1}{1.67\times 10^{-27}}=4.79\times 10^5m/sec$

Now we have to find the momentum

Momentum is equal to product of mass and velocity

So $P=mv=1.67\times 10^{-27}\times 4.79\times 10^5=8\times 10^{-22}kgm/sec$

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