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A proton from cosmic rays enters the earth'smagnetic field in a direction perpendicular to thefield. If the velocity of proton is 2x10' m/s andB 1.6 x 106 wb/m2, find the force exerted on theproton by the magnetic field.(Charge on a proton (e) 1.6 x 10 1°c)(Ans : 5.12 x 10 18N) |
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Answer» we know that Force = qVB(sin 90= 1))so after putting all the value= 1.6*2*2*10⁻¹⁸N5.12*10⁻¹⁸NAnswer Thanks |
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