A proton of velocity $10^{6} m / s$ is moving perpendicular to a uniform magnetic field of $10 kilogauss$ . The sideways force acting on the proton is [[1]] $\times 10^{- 13} N$ .

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Detailed Solution

A proton of velocity

10^{6} m / s

is moving perpendicular to a uniform magnetic field of

10 kilogauss

. The sideways force acting on the proton is

1.6 \times 10^{- 13} N

.
Given,
Magnetic field,

B = 10 kilogauss

Converting it to gauss,

B = 10^{4} G = 1 T

Where, T = tesla is the SI unit of magnetic field
Velocity of the proton,

v = 10^{6} m / s

We know that charge on proton,

q = 1.6 \times 10^{- 19} C

Magnetic field is the region in which electric charge experience magnetic influence. Due to the magnetic field, the moving charges experience force acting on them. The direction of force is perpendicular to the velocity and to the magnetic field.
The force is given by