Question

please answer this question please please​

Answer 1

Assume the current flows through the cylinder outside the plane, towards to us.

By Ampere's Circuital Law, the magnetic field due to the large cylinder at Q is given by,

$\sf{\longrightarrow B_1\cdot2\pi(2R)=\mu_0I}$

Here $\sf{I=\pi R^2J.}$ Then,

$\sf{\longrightarrow B_1\cdot4\pi R=\mu_0\pi R^2J}$

$\sf{\longrightarrow B_1=\dfrac{\mu_0JR}{4}}$

This field acts along y direction. So in vector form,

$\longrightarrow\vec{\sf{B_1}}=\sf{\dfrac{\mu_0JR}{4}\ \hat j}$

The magnetic field due to the upper cavity is acting at Q as the following.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\put(0,0){\line(1,0){60}}\put(0,0){\line(0,1){15}}\put(60,0){\line(-4,1){60}}\qbezier(48,0)(49,2)(50,2.5)\put(60,0){\vector(1,4){5}}\multiput(60,0)(0,5.72){4}{\line(0,1){2.86}}\qbezier(60,8.4)(61,8)(61.4,6.5)\multiput(44,0.5)(16.5,10){2}{ \theta }\put(0,0.2){\framebox(1.5,1.5)}\put(60,0){\circle*{1}}\footnotesize\put(-5,6){ \sf{\dfrac{R}{2}} }\put(27.5,-4){\sf{2R}}\put(27.5,12){ \sf{\dfrac{R}{2}\sqrt{17}} }\put(66,20){ \vec{\sf{B_2}} }\put(61,-4){\sf{Q}}\put(-3,-4){\sf{O}}\put(-3,17){\sf{M}}\end{picture}$

O is center of the large cylinder and M is center of the upper cavity.

By Ampere's Circuital Law, this magnetic field is given by,

$\sf{\longrightarrow B_2\cdot2\pi\left(\dfrac{R}{2}\sqrt{17}\right)=\mu_0I'}$

where $\sf{I'}$ is the current through upper or lower cavity given by,

• $\sf{I'=\pi\left(\dfrac{R}{2}\right)^2J=\dfrac{\pi R^2J}{4}}$

So,

$\sf{\longrightarrow B_2\pi R\sqrt{17}=\dfrac{\mu_0\pi R^2J}{4}}$

$\sf{\longrightarrow B_2=\dfrac{\mu_0JR}{4\sqrt{17}}}$

In vector form, from the figure,

$\longrightarrow\vec{\sf{B_2}}=\sf{\dfrac{\mu_0JR}{4\sqrt{17}}\left(\sin\theta\,\hat i+\cos\theta\,\hat j\right)}$

The magnetic field due to the lower cavity is acting at Q as the following.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\put(0,0){\line(1,0){60}}\put(0,0){\line(0,-1){15}}\put(60,0){\line(-4,-1){60}}\qbezier(48,0)(49,-2)(50,-2.5)\put(60,0){\vector(-1,4){5}}\multiput(60,0)(0,5.72){4}{\line(0,1){2.86}}\qbezier(59.9,8.4)(58.9,8)(58.5,6.5)\multiput(44,-3.5)(13.5,14){2}{ \theta }\put(0,-1.5){\framebox(1.5,1.5)}\put(60,0){\circle*{1}}\footnotesize\put(-5,-10){ \sf{\dfrac{R}{2}} }\put(27.5,2){\sf{2R}}\put(27.5,-14){ \sf{\dfrac{R}{2}\sqrt{17}} }\put(50,20){ \vec{\sf{B_3}} }\put(61,-4){\sf{Q}}\put(-3,2){\sf{O}}\put(-3,-19){\sf{N}}\end{picture}$

N is center of the lower cavity.

The field $\vec{\sf{B_3}}$ has same magnitude of $\vec{\sf{B_2}}.$ From the figure,

$\longrightarrow\vec{\sf{B_3}}=\sf{\dfrac{\mu_0JR}{4\sqrt{17}}\left(-\sin\theta\,\hat i+\cos\theta\,\hat j\right)}$

Now, the net magnetic field at Q is,

$\longrightarrow\vec{\sf{B}}=\vec{\sf{B_1}}-\vec{\sf{B_2}}-\vec{\sf{B_3}}$

$\longrightarrow\vec{\sf{B}}=\sf{\dfrac{\mu_0JR}{4}\ \hat j-\dfrac{\mu_0JR}{4\sqrt{17}}\left(\sin\theta\,\hat i+\cos\theta\,\hat j-\sin\theta\,\hat i+\cos\theta\,\hat j\right)}$

$\longrightarrow\vec{\sf{B}}=\sf{\dfrac{\mu_0JR}{4}\ \hat j-\dfrac{2\mu_0JR}{4\sqrt{17}}\cos\theta\,\hat j}$

From the figure,

$\longrightarrow\vec{\sf{B}}=\sf{\dfrac{\mu_0JR}{4}\ \hat j-\dfrac{\mu_0JR}{2\sqrt{17}}\cdot\dfrac{2R}{\left(\dfrac{R}{2}\sqrt{17}\right)}\,\hat j}$

$\longrightarrow\vec{\sf{B}}=\sf{\dfrac{\mu_0JR}{4}\ \hat j-\dfrac{\mu_0JR}{2\sqrt{17}}\cdot\dfrac{4}{\sqrt{17}}\,\hat j}$

$\longrightarrow\vec{\sf{B}}=\sf{\dfrac{\mu_0JR}{4}\ \hat j-\dfrac{2\mu_0JR}{17}\,\hat j}$

$\longrightarrow\vec{\sf{B}}=\sf{\dfrac{9}{68}\,\mu_0JR\,\hat j}$

Hence,

$\sf{\longrightarrow\underline{\underline{\alpha=9}}}$