Question

A proton enter a magnetic field of flux density 1.5 with a speed of 2×10^7 at angle of 3o degree with the field .The force on proton will be

Answer 1

here is the answer

2.4 * 10 raise to power -12 N The magnitude of the magnetic force on a charg particle is F = |q| v B sin θ Here q = charge on a proton = 1.602 x 10−19 C v = 2 x 107 m/s θ = 30° B = 1.5 T Substituting all these values in the expression for F, we get F = 1.602 x 10−19 C x 2 x 107 m/s x 1.5 T sin 30° = 2.403 x 10−12 N

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