Question

A $250 -$ Turn rectangular coil of length $2.1cm$ and width $1.25cm$ carries a current of $85\mu A$ and is subjected to a magnetic field of strength $0.85T$. Work done for rotating the coil ${180^ \circ }$ against the torque is:
$\left( a \right)$ $1.15\mu J$
$\left( b \right)$ $9.1\mu J$
$\left( c \right)$ $4.55\mu J$
$\left( d \right)$ $23\mu J$

Answer 1

.Hint. In this question, we have been given the number of turns in the coil and its length, width, and also the current and the magnetic field. So by using the formula for the work done by rotation, we will be able to find the work done by it.
Formula used:
Work is done for rotating,
$W = MB\left( {\cos {\theta _1} - \cos {\theta _2}} \right)$
Here,
$W$, will be the work done
$M$, will be the magnetic moment and
$B$, will be the magnetic field
And
Magnetic moment,
$M = NIA$
Here,
$N$, will be the number of turns in the coil
$I$, will be the current
$A$, will be the area

.Complete Step by Step Solution. So first of all we will see the values which are given to us.
Number of turns $N = 250$
Length $l = 2.1cm$
Width $d = 1.25cm$
Current $I = 85\mu A$
Magnetic field $B = 0.85T$
So now as we know,
Work is done for rotating,
$W = MB\left( {\cos {\theta _1} - \cos {\theta _2}} \right)$
So when it is rotated by the angle of ${180^ \circ }$
$\Rightarrow W = 2MB$
And we know already $M = NIA$
Therefore, putting this equation in the above equation, we get
$\Rightarrow W = 2\left( {NIA} \right)B$
Now putting the values, we get
$\Rightarrow W = 2 \times 250 \times 85 \times {10^{ - 6}}\left[ {2.1 \times 1.25 \times 10 - 4} \right] \times 85 \times {10^{ - 2}}$
On solving the above particular equation, we get
$\Rightarrow W = 0.0000091J$
So it can also be written as
$\Rightarrow W = 9.1\mu J$

.Therefore, $9.1\mu J$ work is done to rotate it by${180^ \circ }$. Thus, the option $B$ is the correct one..

.Note. Work done is a proportion of devouring energy or it is the limit of accomplishing certain work and that additionally implies the equivalent. And one thing we have to remember that net work is done for practical purposes and macro systems we will not account for the internal forces of the body as work is done we will define for the external forces acting on a body. And the work or capacity of internal forces we will measure in the form of energy.