A uniform magnetic field of $3000 \mathrm{G}$ is established along the positive z-direction. A rectangular loop of sides $10 \mathrm{~cm}$ and $5 \mathrm{~cm}$ carries a current of $12 \mathbf{A} .$ What is the torque on the loop in the different cases shown in the figure? What is the force on each case? Which case corresponds to stable equilibrium?

Home » A uniform magnetic field of is established along the positive z-direction. A rectangular loop of sides and carries a current of What is the torque on the loop in the different cases shown in the figure? What is the force on each case? Which case corresponds to stable equilibrium?

A uniform magnetic field of $3000 \mathrm{G}$ is established along the positive z-direction. A rectangular loop of sides $10 \mathrm{~cm}$ and $5 \mathrm{~cm}$ carries a current of $12 \mathbf{A} .$ What is the torque on the loop in the different cases shown in the figure? What is the force on each case? Which case corresponds to stable equilibrium?

CBSE | Class 12 | Moving Charges and Magnetism | Physics

A uniform magnetic field of $3000 \mathrm{G}$ is established along the positive z-direction. A rectangular loop of sides $10 \mathrm{~cm}$ and $5 \mathrm{~cm}$ carries a current of $12 \mathbf{A} .$ What is the torque on the loop in the different cases shown in the figure? What is the force on each case? Which case corresponds to stable equilibrium?

Magnetic field strength is given as $B=3000 \mathrm{G}=0.3 \mathrm{~T}$
Area of the loop will be, $A=10 \times 5=50 \mathrm{~cm}^{2}=50 \times 10^{-4} \mathrm{~m}^{2}$
Current flowing in the loop is, $\mathrm{I}=12 \mathrm{~A}$
(a) Torque, $\vec{\tau}=I \vec{A} \times \vec{B}$
$B$ is directed along the z- axis and $A$ is normal to $y-z$ plane
$\vec{A}=50 \times 10^{-4} \hat{i}$
$\vec{B}=0.3 \hat{k}$
$\vec{\tau}=12 \times\left(50 \times 10^{-4}\right) \hat{i} \times 0.3 \hat{k}$
$=-1.8 \times 10^{-2} \hat{j} N m$
Along the negative $\mathrm{y}$ -direction, the torque is $1.8 \times 10^{-2} \mathrm{~N} \mathrm{~m}$ . The net force on the loop is zero because the opposing forces operating on the loop’s two ends cancel each other out.
(b) This is the same as (a). As a result, the torque is $1.8 \times 10^{-2} \mathrm{~N} \mathrm{~m}$ along the negative $y$ -direction. The net force is equal to zero.