Question

Q1.A wire of fixed length is bent in the shape of a circle of single
turn. If a current / flows in the circular loop then the magnetic
field at the centre of the loop was found to be equal to B. Now
the number of turns in the loop are increased to 3 and the
current is reduced by a factor of 3. The magnetic field produced
at the centre of the loop will be

I.b/3
ii.b/4
iii.9b
iv.3b

Answer 1

Explanation:

Given:

A wire of fixed length is bent in the shape of a circle of single turn. If a current / flows in the circular loop then the magnetic field at the centre of the loop was found to be equal to B. Now

the number of turns in the loop are increased to 3 and the current is reduced by a factor of 3.

To find:

New magnetic field intensity at centre?

Calculation:

First of all, initial magnetic field intensity:

$\rm \: B = \dfrac{ \mu_{0}i}{2r}$

Now, when 3 turns are incorporated, net radius of coil will be :

$</p><p>\rm \: 2\pi r = 3 \times (2\pi r_{2} )</p><p> </p><p>$

$\rm \implies \: r_{2} = \dfrac{r}{3} </p><p>$

Now, final magnetic field intensity:

$\rm \: B_{2} = N \times \dfrac{ \mu_{0} \: i_{2} }{2 \: r_{2} }$

$\rm \implies \: B_{2} = 3 \times \dfrac{ \mu_{0} \: (\frac{i}{3} ) }{2 \: (\frac{r}{3} ) }$

$\rm \implies \: B_{2} = 3 \times \dfrac{ \mu_{0} i }{2 r}$

$\rm \implies \: B_{2} = 3 B$

So, new field intensity is 3B.