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Byju's Answer

Standard XII

Physics

Force on a Current Element

The figure sh...

Question

The figure shows a square loop, each side of $10 cm$ in the $x y -$ plane, with its centre at the origin. An infinite wire is at $z = 12 cm$ above $y -$ axis. If the torque on loop due to magnetic force is expressed as $x \times 10^{- 7} N-m$ , fill the value of $x$ .

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Solution

Force on a current carrying conductor,

$\to F = i (\to l \times \to B)$

So, $F = i l B sin θ$

For, sides $A B$ and $C D$ , $θ = 0^{\circ}$ and $180^{\circ}$ respectively.

$∴ F = 0$ on these sides.

For sides $D A$ and $B C$ , $θ = 90^{\circ}$

$∴ F = i l B$ on these sides

From figure,

$d = \sqrt{5^{2} + 12^{2}} = 13 cm$

So,

${\to F}_{D A} = 78 \times 0.01 \times \frac{μ_{o} i}{2 π d} = 78 \times 0.01 \times 10^{- 7} \times \frac{65}{0.13}$

$\Rightarrow {\to F}_{D A} = 3.9 \times 10^{- 5} N$

$∴ τ_{D A} = r F sin θ = 0.05 \times 3.9 \times 10^{- 5} \times sin (180^{\circ} - θ)$

$\Rightarrow τ_{D A} = 0.05 \times 3.9 \times 10^{- 5} \times sin θ$

$\Rightarrow τ_{D A} = 0.05 \times 3.9 \times 10^{- 5} \times \frac{12}{13} = 1.8 \times 10^{- 6} N-m (clockwise)$

Similarly,

$τ_{B C} = 1.8 \times 10^{- 6} N-m (clockwise)$

So, net torque,

$τ_{n e t} = 2 \times 1.8 \times 10^{- 6} = 3.6 \times 10^{- 6} = 36 \times 10^{- 7} N-m (clockwise)$

$∴ x = 36$

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The figure shows a square loop, each side of 10 cm in the xy−plane, with its centre at the origin. An infinite wire is at z=12 cm above y−axis. If the torque on loop due to magnetic force is expressed as x×10−7 N-m, fill the value of x.

The figure shows a square loop, each side of $10 cm$ in the $x y -$ plane, with its centre at the origin. An infinite wire is at $z = 12 cm$ above $y -$ axis. If the torque on loop due to magnetic force is expressed as $x \times 10^{- 7} N-m$ , fill the value of $x$ .