Question

Rectangular coil of $100$ turns has a length of $5cm$ and width $4cm$ . It is placed with its plane parallel to the plane of uniform magnetic field and a current of $2$ Ampere is sent through the coil. Find the magnitude of magnetic field B, if torque acting on the rectangular coil is $0.2N{m^{ - 1}}$ .

Answer 1

Hint: When a rectangular coil is placed with its plane parallel to the plane of magnetic field then the direction of Area vector of plane is perpendicular to the direction of magnetic field and we will use torque formula which is given as $\vec \tau = ni\vec A \times \vec B$ to find the magnitude of magnetic field.

Complete answer:
As we have given that,
Length of rectangular coil is $ = 5cm$
Or, $L = 0.05m$
Width of coil is $ = 4cm$
Or, $W = 0.04m$
Area of rectangular coil is $ = L \times W$
$A = 0.05 \times 0.04{m^2}$
$A = 0.002{m^2}$
And magnitude of torque acting on it is given by $\tau = 0.2N{m^{ - 1}}$
Current in the coil is given by $i = 2A$
Number of turns in the coil is $n = 100$
Using, torque formula $\vec \tau = ni\vec A \times \vec B$
As Magnetic field and area vector ma angle of ${90^ \circ }$ so,
$\tau = niAB\sin {90^ \circ }$
$0.2 = 100 \times 2 \times 0.002 \times B$
$B = 0.5Tesla$
Hence, the magnitude of the magnetic field in which the rectangular coil is placed is $B = 0.5Tesla$ .

Note: It should be remembered that the direction of area vector of any surface is normal to the surface and hence when two planes are parallel then the area vector of one plane is always perpendicular to the other plane surface. The basic conversions used should be remembered which are $1m = 100cm$ and the magnitude of $\sin {90^ \circ } = 1$ . The SI unit of magnetic field is Tesla denoted by T.