A proton enters a magnetic field of flux dens | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

In a slide caliper, \((m+1)\) number of vernier divisions is equal to \(m\) number of smallest main scale divisions. If \(d\) unit is the magnitude of the smallest main scale division, then the magnitude of the vernicr constant is

$A$ plane $\pi$ makes intercepts $3$ and $4$ respectively on $Z$-axis and $X$-axis. If $\pi$ is parallel to $Y$-axis,then its equation is:

Arrange $Ce^{+3}, La^{+3}, Pm^{+3}$ and $Yb^{+3}$ in increasing order of their ionic radii.

Given that,$a \alpha^2+2 b \alpha+c \neq 0$ and that the system of equations
$\begin{aligned} & (a \alpha+b) x+a y+b z=0 \\ & (b \alpha+c) x+b y+c z=0 \\ & (a \alpha+b) y+(b \alpha+c) z=0\end{aligned}$
has a non-trivial solution,then $a, b$ and $c$ lie in

The polymer having an amide linkage is

Vedclass Test Series

Exam Paper Generator

Online Exam Module