magnetic field

Electricity and Magnetism Level pending

Assume in the plane of the earth's magnetic equator the planet's field is uniform with the value $25\times10^{6}Tesla$ northward perpendicular to this plane, everywhere inside a radius of $100Megametre$ . Also assume the earth's field is zero outside the circle. A cosmic ray proton travelling at one-tenth of the speed of light is heading directly outward the center of the earth in the plane of the magnetic equator. Find the radius of curvature of the path it follows when it enters the region of the planet's assumed field.

288Km

36Km

12.5Km

123Km

1 solution

Samuel Ayinde
Mar 10, 2015

using, $F=\frac{mv^{2}}{r}=qVB$ ,

therefore, $\frac{mv^{2}}{r}=qVB$ ,

therefore, $\frac{mv}{r}=qB$ ,

therefore, $r=\frac{mv}{qB}$ ,

$r= \displaystyle\frac{1.673\times10^{-27} \times 3\times10^{7}}{1.6\times10^{-19} \times 25\times10^{-6}}$

$r= 12.5Km$

magnetic field

1 solution

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