
A particle with specific charge s is fired with speed v towards the wall at a distance d perpendicular to the wall. What minimum magnetic field will exist in this region for a particle not to hit the wall?
\[\begin{align}
& A)\dfrac{v}{sd} \\
& B)\dfrac{2v}{sd} \\
& C)\dfrac{v}{2sd} \\
& d)\dfrac{v}{4sd} \\
\end{align}\]
Answer
536.1k+ views
Hint: When a charged particle is moving in a magnetic field then it experiences a magnetic force. When a charge particle moves perpendicular to a magnetic field the maximum Lorentz force acts on a charge particle and it moves in a circular path of defined radius and it requires centripetal force to complete its circular motion.
Complete answer:
The charge particle of specific charge s is thrown with speed v towards the wall at perpendicular distance d from the wall and it will not hit the wall if it will just miss the wall and this could only happen when the charge particle takes circular path of radius d.
Since we know that the charge particle follows a circular path only when it is thrown with some velocity perpendicular to the magnetic field.
So we can say that magnetic fields will exist in the perpendicular direction of velocity of charge particles. So a charged particle will move in a circular path just touching the wall and for circular motion centripetal force is required which is provided by some other force here this force is magnetic Lorentz force.
So we can say that Centripetal force is equal to Magnetic Lorentz force
\[{{F}_{centripetal}}={{F}_{magnetic}}\]
On solving we get,
\[\dfrac{m{{v}^{2}}}{r}=qvB\]
\[\Rightarrow B=\dfrac{mv}{qr}\]
\[\Rightarrow B=\dfrac{v}{\dfrac{qr}{m}}\]
Specific charge in question is s which means charge to mass ratio is s.
\[\Rightarrow s=\dfrac{q}{m}\]
Put the value of s in above equation we get,
\[B=\dfrac{v}{sd}\]
\[\therefore \]So Correct Option is A.
Note:
Centripetal force is always provided by some other source of force. Here Centripetal force is provided by Magnetic Lorentz force and we know that centripetal force will only be activated when a charged particle moves in the magnetic field that is always perpendicular to that magnetic field.
Complete answer:
The charge particle of specific charge s is thrown with speed v towards the wall at perpendicular distance d from the wall and it will not hit the wall if it will just miss the wall and this could only happen when the charge particle takes circular path of radius d.
Since we know that the charge particle follows a circular path only when it is thrown with some velocity perpendicular to the magnetic field.
So we can say that magnetic fields will exist in the perpendicular direction of velocity of charge particles. So a charged particle will move in a circular path just touching the wall and for circular motion centripetal force is required which is provided by some other force here this force is magnetic Lorentz force.
So we can say that Centripetal force is equal to Magnetic Lorentz force
\[{{F}_{centripetal}}={{F}_{magnetic}}\]
On solving we get,
\[\dfrac{m{{v}^{2}}}{r}=qvB\]
\[\Rightarrow B=\dfrac{mv}{qr}\]
\[\Rightarrow B=\dfrac{v}{\dfrac{qr}{m}}\]
Specific charge in question is s which means charge to mass ratio is s.
\[\Rightarrow s=\dfrac{q}{m}\]
Put the value of s in above equation we get,
\[B=\dfrac{v}{sd}\]
\[\therefore \]So Correct Option is A.
Note:
Centripetal force is always provided by some other source of force. Here Centripetal force is provided by Magnetic Lorentz force and we know that centripetal force will only be activated when a charged particle moves in the magnetic field that is always perpendicular to that magnetic field.
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