Question

Do magnets affect electric currents?

Answer 1

Explanation:

Magnets affecting electric currents :::

Electric and magnetic fields are unified into an electromagnetic field. The electromagnetic field is described in spacetime by the electromagnetic tensor.

♥ Different observers will see different combinations of the electric and magnetic components.

♥ When the charge is moving wrt to the magnetic field, there will be a force that will act upon the charge.

♥ An observer stationary wrt to the magnetic field will see only the magnetic field component that acts upon the charge. But from the charge's frame of reference, along with the magnetic field there is also an electric field component.

♥ So the complete equation for the Lorentz force that takes into account all the inertial frames of reference is:

F=qE+qv×B

hope it helps u......❣️

Answer 2

Explanation:

Formula of Potassium Chloride is \bold{KCl.}KCl.{\mathcal{\purple{Q}\green{U}\pink{E}\blue{S}\purple{T}\green{I}\pink{O}\red{N}\blue{!}}}QUESTION!

Lét a, b ānd c be thrêe Intégral válues süch thāt :

a, b, c > 0 & ab + bc + ca ≈ 1 .

Thén prôvë thãt -

\sf{ \bold { \sqrt[2]{a^3+a} + \sqrt[2]{b^3 + b } + \sqrt{ c^3 + c} \geqslant 2 \sqrt[2]{a + b + c } }}2a3+a+2b3+b+c3+c⩾22a+b+c

{\mathcal{\purple{A}\green{N}\pink{S}\blue{W}\purple{E}\green{R}\pink{!}}}ANSWER!

→ Since a , b , c are in AP , we have

› 2b = (a + c) ..............(i)

Now,

\begin{gathered}\sf \frac{1}{ \sqrt{b} + \sqrt{c}} , \frac{1}{ \sqrt{c} + \sqrt{a} } , \frac{1}{ \sqrt{a} + \sqrt{b} } \: are \: in \: AP. \\\end{gathered}b+c1,c+a1,a+b1areinAP.

will be in AP

\begin{gathered}\sf if \: \frac{1}{( \sqrt{c} + \sqrt{a} )} - \frac{1}{ (\sqrt{b} + \sqrt{c} )} \\ = \frac{1}{ (\sqrt{a} + \sqrt{b} )} - \frac{1}{( \sqrt{c} + \sqrt{a} )} \\ \\ \sf \: i.e. \: if \: \frac{ (\sqrt{ b} - \sqrt{a}) }{( \sqrt{c} + \sqrt{a} )( \sqrt{b} + \sqrt{c} )} \\ = \frac{( \sqrt{ c} - \sqrt{b} )}{( \sqrt{a} + \sqrt{b} )( \sqrt{c} + \sqrt{a) } } \\ \\ \sf \: i.e. \: if \frac{( \sqrt{b} - \sqrt{a} )}{( \sqrt{b} + \sqrt{c} )} = \frac{( \sqrt{c} - \sqrt{b}) }{( \sqrt{a} + \sqrt{b}) } \\ \\ \sf i.e \: if \: b - a = c - b \\ \\ \sf \: i.e \: if \: 2b \: = a + c \: from \: (i)\end{gathered}if(c+a)1−(b+c)1=(a+b)1−(c+a)1i.e.if(c+a)(b+c)(b−a)=(a+b)(c+a)(c−b)i.e.if(b+c)(b−a)=(a+b)(c−b)i.eifb−a=c−bi.eif2b=a+cfrom(i)

•°• a,b,c are in AP

\begin{gathered}\sf \frac{1}{ \sqrt{b} + \sqrt{c}} , \frac{1}{ \sqrt{c} + \sqrt{a} } , \frac{1}{ \sqrt{a} + \sqrt{b} } \: are \: in \: AP. \\\end{gathered}b+c1