A square loop of side $a$ hangs from an insulating hanger of a spring balance. The magnetic field of strength $B$ present only at its lower edge which carries a current $I$ , as shown in the figure. Find the change in the reading of the spring balance if the direction of current is reversed-

I a B

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 I a B

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{I a B}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{3}{2} I a B

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $2 I a B$
As, spring balance reads tension in the spring.

Initially net tension in the spring is,

$F_{1} = m g + I a B$ (downwards)

When direction of current is reversed, now tension in the spring is,

$F_{2} = m g - I a B$ (downwards)

$Δ F = F_{1} - F_{2} = 2 I a B$

 Hence, $(B)$ is the correct answer.

Suggest Corrections

Similar questions

Q. A square loop of side 'a' hangs from an insulating hanger of spring balance. The magnetic field of strength B occurs only at the lower edge. It carries a current I. Find the change in the reading of the spring balance, if the direction of current is reversed.

Q. A rectangular wire-loop of width a is suspended from the insulated pan of a spring balance, as shown in the figure. A current i exists in the anti-clockwise direction in the loop. A magnetic field B exists in the lower region. Find the change in the tension of the spring if the current in the loop is reversed.

A rectangular wire-loop of width a is suspended from the insulated pan of a spring balance as shown in figure (34 - E5). A current i exists in the anticlock wise direction in the loop. A magnetic field B exists in the lower region. Find the change in the tension of the spring if the current in the loop is reversed.