A small coil C with $N = 200$ turns is mounted on one end of a balance beam and introduced between the poles of an electromagnet as shown in fig. The area of the coil is $S = 1 c m^{2}$ , the length of the right arm of the balance beam is $l = 30 c m$ . When there is no current in the coil the balance is in equilibrium. On passing a current $I = 22 m A$ through the coil, equilibrium is restored by putting an additional weight of mass $m = 60$ mg on the balance pan. Find the magnetic induction field (in terms of $\times 1 0^{- 1} T)$ between the poles of the electromagnet, assuming it to be uniform :

Question

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Answer 1

Torque on the loop due to magnetic field $τ_{m} = N I S B sin 9 0^{\circ}$
Torque due to extra weight added $τ_{g} = (Δ M) g l$
Equating both torques, $N I S B sin 9 0^{\circ} = (Δ M) g l$

$∴ B = \frac{( Δ M ) g l}{N I S} = \frac{( 6 0 \times 1 0 ^{- 6} ) \times 9 . 8 \times ( 3 0 \times 1 0 ^{- 2} )}{2 0 0 \times ( 2 2 \times 1 0 ^{- 3} )} = 0.4 T = 4 \times 1 0^{- 1} T$

Hence, the magnetic field in units of $1 0^{- 1} T$ is 4.

Answer 2

Torque on the loop due to magnetic field

τ_{m} = N I S B sin 9 0^{\circ}

Torque due to extra weight added

τ_{g} = (Δ M) g l

Equating both torques,

N I S B sin 9 0^{\circ} = (Δ M) g l

∴ B = \frac{( Δ M ) g l}{N I S} = \frac{( 6 0 \times 1 0 ^{- 6} ) \times 9 . 8 \times ( 3 0 \times 1 0 ^{- 2} )}{2 0 0 \times ( 2 2 \times 1 0 ^{- 3} )} = 0.4 T = 4 \times 1 0^{- 1} T

Hence, the magnetic field in units of

1 0^{- 1} T

is 4.