A ring of uniform mass density having mass m=100 gm and radius R=50 cm is suspended by two strings PS and PT from the point P on y-axis in vertical plane. The ring is kept parallel to horizontal surface on x-z plane. Ring carries a current i=722amp as shown in the figure. There exist a magnetic field (B= 1 Tesla) along negative x-axis. The tension in the string PS is T1 and in the string PT is T2. Find the ratio T1T2. (It is given that θ=60∘)(Take g=10m/s2)

T1cos60∘+T2cos60∘=mgT1+T2=2mg.....(i)Taking torque about ‘O’(T2cos60∘)R+|μ→×B→|=(T1cos60∘)R......(ii)on solving, we get T1T2=3

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A ring of uniform mass density having mass $m = 100 gm$ and radius $R = 50 cm$ is suspended by two strings PS and PT from the point P on y-axis in vertical plane. The ring is kept parallel to horizontal surface on x-z plane. Ring carries a current $(i = \frac{7}{22} amp)$ as shown in the figure. There exist a magnetic field (B= 1 Tesla) along negative x-axis. The tension in the string PS is $T_{1}$ and in the string PT is $T_{2}$ . Find the ratio $\frac{T_{1}}{T_{2}}$ . (It is given that $θ = 60^{\circ}$ )

$(Take g = 10 m / s^{2})$

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$T_{1} \cos 60^{\circ} + T_{2} \cos 60^{\circ} = mg$

$T_{1} + T_{2} = 2 mg . .... (i)$

Taking torque about ‘O’

$(T_{2} \cos 60^{\circ}) R + | \vec{μ} \times \vec{B} | = (T_{1} \cos 60^{\circ}) R . ..... (ii)$
on solving, we get $\frac{T_{1}}{T_{2}} = 3$

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