A charge in motion

A particle of mass m m and charge q q is projected with v = v 0 i ^ \overrightarrow { v } ={ v }_{ 0 }\hat { i } in a non uniform magnetic field which is given by B = B 0 x k ^ \rightarrow \overrightarrow { B } =-{ B }_{ 0 }x\hat { k } , where B 0 , v 0 { B }_{ 0 }, { v }_{ 0 } are positive constants. Find the maximum x x coordinate of the particle during its motion.

Details and Assumptions

Assume that the particle enters the x y x-y plane at origin with the given velocity.

v 0 m 2 q B 0 \sqrt { \frac { { v }_{ 0 }m }{ 2q{ B }_{ 0 } } } 2 v 0 m q B 0 \sqrt { \frac { {2 v }_{ 0 }m }{ q{ B }_{ 0 } } } v 0 m q B 0 \sqrt { \frac { { v }_{ 0 }m }{ q{ B }_{ 0 } } } 2 v 0 m q B 0 2\sqrt { \frac { { v }_{ 0 }m }{ q{ B }_{ 0 } } }

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2 solutions

Swagat Panda
Jul 16, 2016

Relevant wiki: Lorentz Force Law (Magnetic and Mixed Fields)

The max x coordinate will be reached when the particle has no velocity in x coordinate, i.e v x = 0 and v y = v 0 \text{The max } x \text{ coordinate will be reached when the particle has no velocity in }x \text{ coordinate, i.e }{ v }_{ x }=0 \text{ and } { v }_{ y }={ v }_{ 0 } . The direction of force is determined using Fleming’s left hand rule, which will come out in + y direction. \text{The direction of force is determined using Fleming's left hand rule, which will come out in }+y \text{ direction.} Even if the angle between v and B is 90 degrees, the particle wont undergo circular motion as the magnetic field is not constant. \text{Even if the angle between } v \text{ and } B \text{ is 90 degrees, the particle wont undergo circular motion as the magnetic field is not constant.}

Now, using Lorentz force equation we get \text{ Now, using Lorentz force equation we get } m a y = q v 0 B 0 x c o s θ m a y = q ( v 0 c o s θ ) B 0 x m d v y d t = q B 0 x d x d t m 0 v 0 d v y = q B 0 0 x m a x x d x m v 0 0 = q B 0 x m a x 2 2 x m a x = 2 m v 0 q B 0 m{ a }_{ y }=q{ v }_{ 0 }{ B }_{ 0 }x cos{ \theta }\quad \Rightarrow \quad m{ a }_{ y }=q({ v }_{ 0 }cos{ \theta })B_{ 0 }x\\ m\dfrac { d{ v }_{ y } }{ dt } =q{ B }_{ 0 }x\dfrac { dx }{ dt } \quad \Rightarrow m\int _{ 0 }^{ { v }_{ 0 } }{ d } { v }_{ y }=q{ B }_{ 0 }\displaystyle\int _{ 0 }^{ { x }_{ max } }{ xdx } \\ m{ v }_{ 0 }-0=q{ B }_{ 0 }\dfrac { { x }_{ max }^{ 2 } }{ 2 } \quad \Rightarrow { x }_{ max }=\sqrt { \dfrac { 2m{ v }_{ 0 } }{ q { B }_{ 0 }} }

yes, I liked this problem so i decided to share it here as well.

Swagat Panda - 4 years, 10 months ago

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which batch? 2016 or 2017? which rank did you get in previous jee mains (in vsat)?

A Former Brilliant Member - 4 years, 10 months ago

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2017 batch, I got my best ever rank in mains 6 which i gave offline, about 206. I gave mains 7 online, so i didn't get any rank, which centre do you belong to?

Swagat Panda - 4 years, 10 months ago

Really nice problem. Keep posting!

Swapnil Das - 4 years, 9 months ago

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Thanks Swapnil, this problem was actually given by our sir for our practice :)

Swagat Panda - 4 years, 9 months ago
Abhi Kumbale
Nov 20, 2016

@Swagat Panda Do see this solution.

Abhi Kumbale - 4 years, 6 months ago

Very nice, frankly speaking, I didn't know the radius of curvature formula for any curve which you mentioned prior to reading your solution. You taught me a new handy formula.

When I attempted this question for the very first time, I tried to use the centripetal force formula with Lorentz force formula to find x m a x x_{max} but because I didn't know at that time that this sort of formula exists, I was unsuccessful. In a way, you fulfilled one of my incomplete wishes to solve this question in another way. :)

Swagat Panda - 4 years, 6 months ago

Nice dude …bdw are u studying in fitjee??

Satyam Tripathi - 4 years, 6 months ago

Yes. I'm from Fiitjee Banglore

Abhi Kumbale - 4 years, 6 months ago

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