Question

A coil of the turn area S= 200cm² is mounted in a uniform magnetic field B = 8.10^-4 T. There is a current of 20 A in the coil, which has 25 turns. When the plane of the coil makes an angle of 60⁰ with direction of the field, what is the torque tending to rotate the coil?

Answer 1

To be given in question

Coil area(A) =200$cm^2$

Coil magnetic field(B)=$8\times10^{-4} T$

Current(i)=20Amp

Turns(n)=25

$\theta=60°$

To be asked in question

$\tau =?$

We know that

$M=niA\rightarrow(1)$

$\tau=M\times B$

$\tau=MBsin\theta\rightarrow(2)$

eqution (1)and(2) we can written as

$\tau=niABsin\theta\rightarrow(3)$

eqution (3)put values

$\tau=25\times20\times200\times10^{-4}\times8\times10^{-4}\times sin60°$ $\tau=4\sqrt{3}\times10^{-3}Nmeter$

$\tau=6.9282\times10^{-3} Nmeter$