A proton enters a magnetic field of flux dens | vedclass

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Question

(A) The magnetic force $F$ on a charged particle moving in a magnetic field is given by the formula: $F = qvB \sin 	heta$. Given: Charge of a proton

Vedclass · Accepted Answer

$2.4 	imes 10^{-12} \, N$

Vedclass · Answer

(A) The magnetic force $F$ on a charged particle moving in a magnetic field is given by the formula: $F = qvB \sin 	heta$. Given: Charge of a proton $q = 1.6 	imes 10^{-19} \, C$,Velocity $v = 2 	imes 10^7 \, m/s$,Magnetic flux density $B = 1.5 \, Wb/m^2$,Angle $	heta = 30^\circ$. Substituting these values into the formula: $F = (1.6 	imes 10^{-19}) 	imes (2 	imes 10^7) 	imes 1.5 	imes \sin(30^\circ)$ $F = (1.6 	imes 10^{-19}) 	imes (2 	imes 10^7) 	imes 1.5 	imes 0.5$ $F = 1.6 	imes 10^{-19} 	imes 10^7 	imes 1.5$ $F = 2.4 	imes 10^{-12} \, N$.

$A$ proton enters a magnetic field of flux density $1.5 \, Wb/m^2$ with a velocity of $2 \times 10^7 \, m/s$ at an angle of $30^\circ$ with the field. The force on the proton will be

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