A particle of mass 1 times 10^{-27} kg and c | vedclass

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

Question

(B) Given: mass $m = 1 	imes 10^{-27} \ kg$,charge $q = 1 	imes 10^{-16} \ C$,speed $v = 1000 \ m/s$,angle $	heta = 60^{\circ}$,number of turns $N

Vedclass · Accepted Answer

$1 	imes 10^6$

Vedclass · Answer

(B) Given: mass $m = 1 	imes 10^{-27} \ kg$,charge $q = 1 	imes 10^{-16} \ C$,speed $v = 1000 \ m/s$,angle $	heta = 60^{\circ}$,number of turns $N = 5000$,current $I = 5 \ A$.The velocity components are:$v_{\parallel} = v \cos 60^{\circ} = 1000 	imes 0.5 = 500 \ m/s$ (along the axis)$v_{\perp} = v \sin 60^{\circ} = 1000 	imes \frac{\sqrt{3}}{2} = 500\sqrt{3} \ m/s$ (perpendicular to the axis)The magnetic field inside the solenoid is $B = \mu_0 n I = \mu_0 (N/L) I$.The time taken to travel the length $L$ of the solenoid is $t = \frac{L}{v_{\parallel}} = \frac{L}{500}$.The time period of one revolution in the magnetic field is $T = \frac{2\pi m}{qB}$.The number of revolutions $n'$ is given by $n' = \frac{t}{T} = \frac{L/v_{\parallel}}{2\pi m / qB} = \frac{L \cdot q \cdot B}{v_{\parallel} \cdot 2\pi m}$.Substituting $B = \frac{\mu_0 N I}{L}$:$n' = \frac{L \cdot q \cdot (\mu_0 N I / L)}{v_{\parallel} \cdot 2\pi m} = \frac{q \mu_0 N I}{v_{\parallel} \cdot 2\pi m}$.Plugging in the values:$n' = \frac{10^{-16} 	imes (4\pi 	imes 10^{-7}) 	imes 5000 	imes 5}{500 	imes 2\pi 	imes 10^{-27}}$$n' = \frac{10^{-16} 	imes 4\pi 	imes 10^{-7} 	imes 25000}{1000\pi 	imes 10^{-27}}$$n' = \frac{10^{-16} 	imes 10^{-7} 	imes 10^5}{10^{-27} 	imes 10^2} = \frac{10^{-18}}{10^{-25}} = 10^7 / 10 = 10^6$.Thus,the number of revolutions is $1 	imes 10^6$.

Similar Questions

$A$ long solenoid of $50\, cm$ length having $100$ turns carries a current of $2.5\, A$. The magnetic field at the centre of the solenoid is $...... \times 10^{-5}\, T$. $(\mu_{0} = 4\pi \times 10^{-7}\, T\, m\, A^{-1})$

$A$ toroid core has an inner radius of $0.24 \ m$ and an outer radius of $0.26 \ m$. $A$ current of $10 \ A$ flows through the wire having $2500$ turns around it. Find the magnetic field inside the core of the toroid.

Calculate the axial magnetic field of a finite solenoid.

$A$ current of $5 \ A$ flows through a toroid having a core of mean radius $20 \ cm$. If $4000$ turns of the conducting wire are wound on the core,then the magnetic field inside the core of the toroid is [permeability of free space $= 4 \pi \times 10^{-7} \ T \cdot m/A$]

$A$ current of $2\, A$ flows in a long,straight wire of radius $2\, mm$. The intensity of the magnetic field on the axis of the wire is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module