Rotating Rod In Magnetic Field

Electricity and Magnetism Level 5

Consider a horizontal circular loop of radius r r carrying a current I I . Concentric and coplanar with the loop is a smooth metallic ring of radius a a , containing a metallic rod O A OA of mass m m , as shown. The rod can freely rotate about O O , the center of arrangement. Between O O and the circumference of ring, a resistor load R R (not shown in the figure) is connected, which doesn't obstruct the motion of rod. The rod is given an angular velocity ω 0 \omega_{0} . Find an expression for the maximum angle θ \theta through which the rod rotates, if a < < r a<<r .

Find the value of 2 ∣ ( θ − 16 3 R m ω 0 ( r μ 0 I a ) 2 ) ∣ \displaystyle 2 \Bigg|\bigg(\theta - \frac{16}{3} R m \omega_{0} \bigg(\frac{r}{\mu_{0} I a} \bigg)^2\bigg)\Bigg| for the values given below :

Details and assumptions
The resistances of the rod and ring can be neglected.
m = 1 g m = 1g
R = 1 Ω R = 1 \Omega
ω 0 = 1 r a d / s \omega_{0} = 1 rad/s
r = 1 m r = 1m
a = 1 c m a = 1cm
I = 1000 A I = 1000 A
μ 0 = 4 π × 1 0 − 7 H / m \mu_{0} = 4 \pi \times 10^{-7} H/m


The answer is 10132.12.

View wiki
Jatin Yadav by jatin yadav , 23, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Discussions for this problem are now closed

Jatin Yadav
Apr 9, 2014

This is going to be one of the lengthiest solutions you have ever seen on brilliant.

I would be repeatedly using ( 1 + α ) n ≈ 1 + n α (1+\alpha)^n \approx 1+n \alpha for α < < 1 \alpha <<1

Consider a small element subtending an angle d θ d \theta at center as shown. Due to all such elements, magnetic field is in same direction and we can directly sum their magnitudes.

Using cosine formula in △ O P Q \triangle OPQ ,

y 2 = r 2 + x 2 − 2 r x cos ⁡ θ \displaystyle y^2 = r^2+x^2-2rx \cos \theta

cos ⁡ ϕ = r 2 + y 2 − x 2 2 r y = r 2 + r 2 + x 2 − 2 r x cos ⁡ θ − x 2 2 r y = r − x cos ⁡ θ y \displaystyle \cos \phi = \frac{r^2+y^2-x^2}{2ry} = \frac{r^2+ r^2 + x^2-2rx \cos \theta - x^2}{2ry} = \frac{r-x \cos \theta}{y}

Now, using Biot savart law,

d B = μ 0 I r d θ cos ⁡ ϕ 4 π y 2 \displaystyle dB = \frac{\mu_{0}I r d\theta \cos \phi}{4 \pi y^2}

d B = μ 0 I r ( r − x cos ⁡ θ ) 4 π ( r 2 + x 2 − 2 r x cos ⁡ θ ) 3 / 2 d θ \displaystyle dB = \frac{\mu_{0} I r(r- x \cos \theta)}{4 \pi (r^2+x^2-2rx \cos \theta)^{3/2}} d \theta

B = μ 0 I r 4 π ∫ 0 2 π r − x cos ⁡ θ ( r 2 + x 2 − 2 r x cos ⁡ θ ) 3 / 2 d θ \displaystyle B = \displaystyle \frac{\mu_{0} I r}{4 \pi} \int_{0}^{2 \pi} \frac{r- x \cos \theta}{(r^2+x^2-2rx \cos \theta)^{3/2}} d \theta

Using approximation for x < < r x <<r , we can show that

B ≈ μ 0 I 4 π r ∫ 0 2 π ( 1 − x r cos ⁡ θ ) ( 1 + 3 x r cos ⁡ θ − 3 2 x 2 r 2 + 15 2 cos ⁡ 2 θ ) d θ B \approx \displaystyle \frac{\mu_{0} I }{4 \pi r} \int_{0}^{2 \pi} \bigg(1 -\frac{x}{r} \cos \theta\bigg) \bigg( 1+ 3 \frac{x}{r} \cos \theta - \frac{3}{2} \frac{x^2}{r^2} + \frac{15}{2} \cos^2 \theta \bigg) d \theta = μ 0 I 2 r ( 1 + 3 2 x 2 r 2 ) \displaystyle \frac{\mu_{0} I}{2 r} \bigg( 1 + \frac{3}{2} \frac{x^2}{r^2} \bigg)

Now, emf induced in the rod is ϵ = ∫ 0 a B ω x d x \epsilon = \displaystyle \int_{0}^{a} B \omega x dx

= μ 0 I ω 2 r ∫ 0 a x ( 1 + 3 2 x 2 r 2 ) d x \displaystyle \frac{\mu_{0} I \omega}{2 r} \int_{0}^{a} x\bigg( 1+ \frac{3}{2} \frac{x^2}{r^2} \bigg) dx

= μ 0 I ω a 2 4 r ( 1 + 3 4 a 2 r 2 ) \displaystyle \frac{\mu_{0} I \omega a^2}{4 r} \bigg(1+ \frac{3}{4} \frac{a^2}{r^2}\bigg)

Now, the current induced is i = ϵ R = μ 0 I ω a 2 4 R r ( 1 + 3 4 a 2 r 2 ) i = \displaystyle \frac{\epsilon}{R} = \frac{\mu_{0} I \omega a^2}{4 R r} \bigg(1 + \frac{3}{4} \frac{a^2}{r^2}\bigg)

Now, the torque due to magnetic field on the rod is τ = ∫ 0 a i B x d x \tau = \displaystyle \int_{0}^{a} i B x dx

= μ 0 I i a 2 4 r ( 1 + 3 4 a 2 r 2 ) \displaystyle \frac{\mu_{0} I ia^2}{4r} \bigg( 1+ \frac{3}{4} \frac{a^2}{r^2}\bigg)

= μ 0 2 I 2 a 4 ω 16 r 2 R ( 1 + 3 4 a 2 r 2 ) 2 \displaystyle \frac{\mu_{0}^2 I^2 a^4 \omega}{16 r^2 R} \bigg(1+ \frac{3}{4} \frac{a^2}{r^2} \bigg)^2

⇒ τ ≈ μ 0 2 I 2 a 4 ω 16 r 2 R ( 1 + 3 2 a 2 r 2 ) \displaystyle \Rightarrow \tau \approx \frac{\mu_{0}^2 I^2 a^4 \omega}{16 r^2 R} \bigg(1 + \frac{3}{2} \frac{a^2}{r^2}\bigg)

Now, Clearly, τ \tau is decreasing angular velocity. Hence,

− m a 2 3 ω d ω d θ = μ 0 2 I 2 a 4 ω 16 r 2 R ( 1 + 3 2 a 2 r 2 ) \displaystyle - \frac{m a^2}{3} \omega \frac{d \omega}{d \theta} = \frac{\mu_{0}^2 I^2 a^4 \omega}{16 r^2 R} \bigg(1 + \frac{3}{2} \frac{a^2}{r^2}\bigg)

⇒ d ω = − 3 μ 0 2 I 2 a 2 16 m r 2 R ( 1 + 3 2 a 2 r 2 ) d θ \Rightarrow \displaystyle d \omega = - \frac{3 \mu_{0}^2 I^2 a^2}{16m r^2 R} \bigg(1 + \frac{ 3}{2} \frac{a^2}{r^2}\bigg) d \theta

⇒ ∫ 0 θ d θ = − ∫ ω 0 0 16 m R 3 ( r μ 0 I a ) 2 ( 1 − 3 2 a 2 r 2 ) d ω \Rightarrow \displaystyle \int_{0}^{\theta} d \theta = - \int_{\omega_{0}}^{0} \frac{16 mR}{3} \bigg(\frac{r}{\mu_{0} I a}\bigg)^2 \bigg( 1 - \frac{3}{2} \frac{a^2}{r^2} \bigg) d \omega

⇒ θ ≈ 16 3 R m ω 0 ( r μ 0 I a ) 2 ( 1 − 3 2 a 2 r 2 ) \Rightarrow \large \boxed{\boxed{\displaystyle \theta \approx \frac{16}{3} R m \omega_{0} \bigg(\frac{r}{\mu_{0} I a} \bigg)^2 \bigg(1 - \frac{ 3}{2} \frac{a^2}{r^2}\bigg)}}

Hence, θ − 16 3 R m ω 0 ( r μ 0 I a ) 2 = − 8 R m ω 0 ( μ 0 I ) 2 \displaystyle \theta - \frac{16}{3} R m \omega_{0} \bigg(\frac{r}{\mu_{0} I a} \bigg)^2 = \large \boxed{\displaystyle \frac{- 8 Rm \omega_{0}}{(\mu_{0} I)^2}}

Put values to get answer as 11032.12 11032.12

8 Helpful 0 Interesting 0 Brilliant 1 Confused

Good solution, but I think this solution of mine is a bit longer than yours.

Anish Puthuraya - 7 years, 2 months ago

When I started writing the solution, I thought that I would be explaining everything, but couldn't. I didn't write everything here in the solution like you did there. Almost everything above is mathematical. There are less statements. This one is surely a lengthier problem. Making this was way more fun(and hard) than making that. Btw your solution there is admirable!

jatin yadav - 7 years, 1 month ago

Thanks

Anish Puthuraya - 7 years, 1 month ago

Sorry, but I think there is a problem with the approximation you made:

( 1 − 2 x r cos ⁡ θ + x 2 r 2 ) − 3 / 2 ≈ ( 1 + 3 x r cos ⁡ θ − 3 2 x 2 r 2 cos ⁡ 2 θ ) (1-2\dfrac{x}{r}\cos \theta+\dfrac{x^2}{r^2})^{-3/2} \approx (1+3\dfrac{x}{r}\cos\theta-\dfrac{3}{2}\dfrac{x^2}{r^2}\cos^2\theta)

where you use ( 1 + x ) n ≈ 1 + x n (1+x)^n \approx 1+xn with x < < 1 x<<1

but you made the approximation to the power of two x 2 r 2 \dfrac{x^2}{r^2} , so it should be:

( 1 + x ) n ≈ 1 + n x + n ( n + 1 ) 2 x 2 (1+x)^n \approx 1+nx+\dfrac{n(n+1)}{2}x^2 .

Therefore, the approximation should be:

( 1 − 2 x r cos ⁡ θ + x 2 r 2 ) − 3 / 2 ≈ ( 1 + 3 x r cos ⁡ θ − 3 2 x 2 r 2 + 15 2 x 2 r 2 cos ⁡ 2 θ ) (1-2\dfrac{x}{r}\cos \theta+\dfrac{x^2}{r^2})^{-3/2} \approx (1+3\dfrac{x}{r}\cos\theta-\dfrac{3}{2}\dfrac{x^2}{r^2}+\dfrac{15}{2}\dfrac{x^2}{r^2}\cos^2\theta)

Đinh Ngọc Hải - 7 years, 1 month ago

Yep! you are true . Thanks for pointing that out. That extra cos ⁡ 2 θ \cos^2 \theta with 3 2 x 2 r 2 \dfrac{3}{2} \dfrac{x^2}{r^2} was typo, while missing the 15 2 x 2 r 2 cos ⁡ 2 θ \dfrac{15}{2} \dfrac{x^2}{r^2} \cos^2 \theta term was a mistake. I have edited both the problem and solution.

By the way did you mean ( 1 + x ) n = 1 + n x + n ( n − 1 ) 2 x 2 \displaystyle (1+x)^n = 1 + nx + \frac{n(n-1)}{2} x^2 ?

This gives coefficient of x 2 x^2 for n = − 3 2 n= - \dfrac{3}{2} as − 3 / 2 ( − 5 / 2 ) 2 = 15 8 \dfrac{-3/2 (-5/2)}{2} = \dfrac{15}{8} . Hence, in our case the we get 15 8 ( 2 x r cos ⁡ θ ) 2 = 15 2 x 2 r 2 cos ⁡ 2 θ \dfrac{15}{8} \bigg(2 \dfrac{x}{r} \cos \theta\bigg)^2 = \dfrac{15}{2} \dfrac{x^2}{r^2} \cos^2 \theta

jatin yadav - 7 years, 1 month ago

Yes, sorry, I meant n ( n − 1 ) 2 \dfrac{n(n-1)}{2}

Đinh Ngọc Hải - 7 years, 1 month ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...