A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is suspended in mid-air by a uniform horizontal magnetic field B. The magnitude of B (in tesla) is (assume g=9.8 ms -2 )

Question

A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is suspended in mid-air by a uniform horizontal magnetic field B. The magnitude of B (in tesla) is (assume g=9.8 ms -2 ) (A) 2 (B) 1.5 (C) 0.55 (D) 0.65

Tardigrade Admin · Accepted Answer

Magnetic force on straight wire F= Bil sin θ= Bil sin 90°= Bil For equilibrium of wire in mid-air, F =m g Bil =m g ∴ B=(m g/il)=(200 × 10-3 × 9.8/2 × 1.5) =0.65 T

Q. A straight wire of mass $200 g$ and length $1.5 m$ carries a current of $2 A$ . It is suspended in mid-air by a uniform horizontal magnetic field $B$ . The magnitude of $B$ (in tesla) is (assume $g = 9.8 m s^{- 2}$ )

2

1.5

0.55

0.65

Magnetic force on straight wire
$F =$ Bil $sin θ =$ Bil $sin 9 0^{\circ} =$ Bil
For equilibrium of wire in mid-air,
$F = m g$
$B i l = m g$
$∴ B = \frac{m g}{i l} = \frac{200 \times 1 0 ^{- 3} \times 9.8}{2 \times 1.5}$
$= 0.65 T$

Q. A straight wire of mass 200g and length 1.5m carries a current of 2A. It is suspended in mid-air by a uniform horizontal magnetic field B. The magnitude of B (in tesla) is (assume g=9.8ms−2 )

2

1.5

0.55

0.65

Magnetic force on straight wire F= Bil sinθ= Bil sin90∘= Bil For equilibrium of wire in mid-air, F=mg Bil=mg ∴B=ilmg​=2×1.5200×10−3×9.8​ =0.65T

Q. A straight wire of mass $200 g$ and length $1.5 m$ carries a current of $2 A$ . It is suspended in mid-air by a uniform horizontal magnetic field $B$ . The magnitude of $B$ (in tesla) is (assume $g = 9.8 m s^{- 2}$ )

Magnetic force on straight wire
$F =$ Bil $sin θ =$ Bil $sin 9 0^{\circ} =$ Bil
For equilibrium of wire in mid-air,
$F = m g$
$B i l = m g$
$∴ B = \frac{m g}{i l} = \frac{200 \times 1 0 ^{- 3} \times 9.8}{2 \times 1.5}$
$= 0.65 T$