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A uniform magnetic field of $3000\;G$ is established along the positive z- direction . A rectangular loop of sides $10\;cm$ and $5\;cm$carries a current of $12\;A$. What is the torque on the loop in the case as shown in the figure? What is the force in this case ?

$\begin{array}{1 1} \text{$T=-1.80 \times 10^{-2} \hat j N-m$, along negative y- direction } \\ \text{$T=1.80 \times 10^{-2} \hat j N-m,$ along positive y- direction } \\ \text{$T=-1.80 \times 10^{-2} \hat j N-m,$ along negative z- direction } \\ \text{zero} \end{array} $

1 Answer

Solution :
Given $ B= 0.3 \hat k T$
$A= -50 \times 10^{-4} \hat k m^2$
$I= 12 \;A$
$T= 12(-50 \times 10^{-4} \hat k \times 0.3 \hat k)$
But $\hat k \times \hat k =0$
Answer : zero
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