Magnetics Ques 88
- A uniform constant magnetic field $\mathbf{B}$ is directed at an angle of $45^{\circ}$ to the $x$-axis in $x-y$ plane. $P Q R S$ is rigid square wire frame carrying a steady current $I _0$, with its centre at the origin $O$. At time $t=0$, the frame is at rest in the position shown in the figure with its sides parallel to $x$ and $y$-axes. Each side of the frame is of mass $M$ and length $L$
$(1998,8$ M)
(a) What is the magnitude of torque $\tau$ acting on the frame due to the magnetic field?
(b) Find the angle by which the frame rotates under the action of this torque in a short interval of time $\Delta t$ and the axis about which this rotation occurs ( $\Delta t$ is so short that any variation in the torque during this interval may be neglected). Given : the moment of inertia of the frame about an axis through its centre perpendicular to its plane is $\frac{4}{3} M L^{2}$.
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Answer:
Correct Answer: 88.(a) $|\tau|=I _0 L^{2} B$ (b) $\theta=\frac{3}{4} \frac{I _0 B}{M}(\Delta t)^{2}$
Solution:
Formula:
- Magnetic moment of the loop, $\mathbf{M}=(i A) \hat{\mathbf{k}}=\left(I _0 L^{2}\right) \hat{\mathbf{k}}$
Magnetic field, $\mathbf{B}=\left(B \cos 45^{\circ}\right) \hat{\mathbf{i}}+\left(B \sin 45^{\circ}\right) \hat{\mathbf{j}}$
$ =\frac{B}{\sqrt{2}}(\hat{\mathbf{i}}+\hat{\mathbf{j}}) $
(a) Torque acting on the loop, $\tau=\mathbf{M} \times \mathbf{B}$
$ =\left(I _0 L^{2} \hat{\mathbf{k}}\right) \times [\frac{B}{\sqrt{2}}(\hat{\mathbf{i}}+\hat{\mathbf{j}})] $
$\therefore \tau =\frac{I _0 L^{2} B}{\sqrt{2}}(\hat{\mathbf{j}}-\hat{\mathbf{i}}) $
$\text { or } |\tau| l=I _0 L^{2} B$
(b) Axis of rotation coincides with the torque and since torque is in $\hat{\mathbf{j}}-\hat{\mathbf{i}}$ direction or parallel to $Q S$. Therefore, the loop will rotate about an axis passing through $Q$ and $S$ as shown in the figure.
Angular acceleration, $\alpha=\frac{|\tau|}{I}$
where, $I=$ moment of inertia of loop about $Q S$.
$ I _{Q S}+I _{P R}=I _{Z Z} $
(From theorem of perpendicular axis)
$\text { But } I _{Q S} =I _{P R} $
$\therefore 2 I _{Q S} =I _{Z Z}=\frac{4}{3} M L^{2} $
$ \therefore I _{Q S} =\frac{2}{3} M L^{2} $
$\alpha =\frac{|\tau|}{I}=\frac{I _0 L^{2} B}{2 / 3 M L^{2}}=\frac{3}{2} \frac{I _0 B}{M}$
$\therefore$ Angle by which the frame rotates in time $\Delta t$ is
$ \theta=\frac{1}{2} \alpha(\Delta t)^{2} \text { or } \theta=\frac{3}{4} \frac{I _0 B}{M} \cdot(\Delta t)^{2} $
BETA
