A uniform rigid, square loop of mass ‘m’ and side ‘l’ is free to rotate about an axis in xy plane passing through two corner as shown in figure. A constant current I is flowing in the loop. In space magnetic field B→(x)=B0(1+xl)(i^+k^) is present. Initially loop is held at rest in xy–plane. On releasing its instantaneous angular acceleration will be (space is gravity free space)

Torque on the loop due to magnetic field τ=∫0l(I2B0dy)y2+∫0l(IB0dy)y2 Iα=3IB0l222 16ml2α=3IB0l222⇒α==(9IB02m)

A uniform rigid, square loop of mass ‘m’ and side ‘l’ is free to rotate about an axis in xy plane passing through two corner as shown in figure. A constant current I is flowing in the loop. In space magnetic field B→(x)=B0(1+xl)(i^+k^) is present. Initially loop is held at rest in xy–plane. On releasing its instantaneous angular acceleration will be (space is gravity free space)

Torque on the loop due to magnetic field τ=∫0l(I2B0dy)y2+∫0l(IB0dy)y2 Iα=3IB0l222 16ml2α=3IB0l222⇒α==(9IB02m)

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A uniform rigid, square loop of mass ‘m’ and side ‘ $l$ ’ is free to rotate about an axis in xy plane passing through two corner as shown in figure. A constant current I is flowing in the loop. In space magnetic field $\vec{B} (x) = B_{0} (1 + \frac{x}{l}) (\hat{i} + \hat{k})$ is present. Initially loop is held at rest in xy–plane. On releasing its instantaneous angular acceleration will be (space is gravity free space)

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$\frac{9 I B}{\sqrt{2} m}$

$\frac{5 I B}{2 \sqrt{2} m}$

$\frac{5 \sqrt{2} I B}{2 m}$

$\frac{9 I B}{2 \sqrt{2} m}$

answer is A.

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Detailed Solution

Torque on the loop due to magnetic field
Question Image
$τ = \int_{0}^{l} (I 2 B_{0} d y) \frac{y}{\sqrt{2}} + \int_{0}^{l} (I B_{0} d y) \frac{y}{\sqrt{2}}$
$I α = \frac{3 I B_{0} l^{2}}{2 \sqrt{2}}$
$\frac{1}{6} m l^{2} α = \frac{3 I B_{0} l^{2}}{2 \sqrt{2}} \Rightarrow α = = (\frac{9 I B_{0}}{\sqrt{2} m})$