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Standard XII

Physics

Rotational EMF

A square loop...

Question

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

\frac{I a B}{2}

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\frac{3}{2} I a B

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2 I a B

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I a B

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Solution

The correct option is C $2 I a B$
Force on current carrying wire placed in magnetic field is
$F = I (\to l \times \to B) Initially (F_{1} = m g + I a B$ (down wards)

When the direction of current is reversed
$F_{2} = m g - I a B$ (down ward)

$\Rightarrow Δ F = 2 I a B$

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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

The correct option is C 2 IaBForce on current carrying wire placed in magnetic field is F=I(→l×→B) Initially (F1=mg+IaB (down wards) When the direction of current is reversed F2=mg−IaB (down ward) ⇒ΔF=2IaB

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

The correct option is C $2 I a B$
Force on current carrying wire placed in magnetic field is
$F = I (\to l \times \to B) Initially (F_{1} = m g + I a B$ (down wards)

When the direction of current is reversed
$F_{2} = m g - I a B$ (down ward)

$\Rightarrow Δ F = 2 I a B$