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12. A proton enters a magnetic ficld of 4T intensity with a velocity of 2.5 x 10 ms 1 at anangle of 30° with the field. Find the magnitude of the force on the proton. Charg onthe proton = 1.602 x 10-19 c |
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Answer» The magnitude of the magnetic force on acharg particle isF =|q| v B sinθ Here, q = charge on a proton = 1.602 x 10−19Cv = 2.5 x 10⁶m/sθ = 30°B = 4 TSubstituting all these values in the expression for F, we get:-F =1.602 x 10−19C x2.5 x 10⁶m/s x 4 T× sin30° = 8×10⁻¹³ N |
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