Question

A circular loop ( radius = 0.5 m) carries a current of 3.0 A and has unit normal vector of (2i - j +2k)/3 what is x component of the torque on this loop when it is placed in a uniform magnetic field of (2i-6j) T

Answer 1

Answer:

$T=9.42Nm$

Explanation:

From the question we are told that:

Radius $r= 0.5 m$

Current $I= 3.0 A$

Normal vector $n=\frac{(2i - j +2k)}{3}$

Magnetic field $B= (2i-6j) T$

Generally the equation for Area is mathematically given by

 $A=\pi r^2$

 $A=3.1415 *0.5^2$

 $A=0.7853 m^2$

Generally the equation for Torque is mathematically given by

 $T=A(i'*B)$

Where

 $i'*B= \begin{bmatrix}2&-1&2\\2&-6&0\end{bmatrix}$

 $X\ component\ of\ i'*B= [(-1 * 0)-(2*-6)]$

 $X\ component\ of\ i'*B=12$

Therefore

 $T=0.7853*12$

 $T=9.42Nm$