Question

A wooden cubical block ABCDEFG of mass m and side a is wrapped by a square wire loop of perimeter 4a, carrying current I. The whole system is placed at frictionless horizontal surface in a uniform magnetic field $\overrightarrow{B}=B_0\hat{j}$ as shown in fig. In this situation, normal force between horizontal surface and block passes through a point at a distance x from center. Choose correct statement(s)

• A. The block must not topple if $I<\frac{mg}{a{B}_0}$
• B. The block must not topple if $I<\frac{mg}{2a{B}_0}$
• C. $x=\frac{a}{4}$ if $I=\frac{mg}{2a{B}_0}$
• D. $x=\frac{a}{4}$ if $I=\frac{mg}{4a{B}_0}$

Answer 1

Answer: (B), (D)

$x=\frac{a}{4}$ if $I=\frac{mg}{4a{B}_0}$
The block must not topple if $I<\frac{mg}{2a{B}_0}$

Torque of magnetic field; $\tau =M{B}_0=I{a}^2{B}_0$
For rotational equilibrium: $\tau =mgx\Rightarrow I{a}^2{B}_0=mgx$
$\Rightarrow x=\frac{I{a}^2{B}_0}{mg}$
For the block not to topple:
$x<\frac{a}{2}\Rightarrow I<\frac{mg}{2a{B}_0}$

If $I=\frac{mg}{4a{B}_o}$, substituting value we get

$x=a/4$