Attraction towards magnet

Electricity and Magnetism Level 5

$A:$ In the figure shown $R$ is a fixed conducting fixed ring of negligible resistance and radius $a$ . $PQ$ is a uniform rod of resistance $r$ .It is hinged at the centre of the ring and rotated about this point in clockwise direction with a uniform angular velocity $\omega$ .There is a uniform magnetic field of strength $B$ pointing inwards. $r$ is a stationery resistance.Then the current through $r$ is $m*\dfrac{Ba^2*\omega}{r}$

$B:$ Let the magnetic field at the centre of the current carrying wire arrangement as shown in the figure be $n*\dfrac{\mu_0*i}{\pi*a}$ , where $a$ is the side length of the innermost square.

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Evaluate $m+n$ .

The answer is 2.3605.

1 solution

Rajdeep Brahma
Jul 2, 2018

$A:$ Equivalent circuit:Induced emf $e$ = $\frac{B\omegaa^2}{2}$

4 $\frac{X-e}{r}$ + $\frac{X-0}{r}=0$

$5X=4e$ and so $I=\frac{2B\omegaa^2}{5r}$

$B:$ $B_{due to first loop}=4$ $\frac{\mu_0i}{4\pi\frac{a}{2}}$ $[cos45+cos45]$ = $\frac{2\sqrt{2}\mu_0i}{\pi*a}$

So,net $B=$ $\frac{2\sqrt{2}\mu_0i}{\pia}$ $(1-\frac{1}{2}+.....)$ = $\frac{2\sqrt{2}\mu_0i}{\pia}$ *ln2

So $m=0.4$ and $n=2\sqrt{2}ln2$ and $m+n=2.3605$

Calculated the current flowing(m) wrong Smh :P

Suhas Sheikh - 2 years, 11 months ago

Ohh sorry to hear

rajdeep brahma - 2 years, 11 months ago

Hey.I solved it