A massless square loop, of wire of resistance 10 Ω, supporting a mass of 1 g, hangs vertically with one of its sides in a uniform magnetic field of 103 G, directed outwards in the shaded region. A de voltage V is applied to the loop. For what value of V, the magnetic force will exactly balance the weight of the supporting mass of 1 g ? (If sides of the loop =10 cm , g =10 ms -2 )

Question

A massless square loop, of wire of resistance 10 Ω, supporting a mass of 1 g, hangs vertically with one of its sides in a uniform magnetic field of 103 G, directed outwards in the shaded region. A de voltage V is applied to the loop. For what value of V, the magnetic force will exactly balance the weight of the supporting mass of 1 g ? (If sides of the loop =10 cm , g =10 ms -2 ) (A) (1/10) V (B) 1 V (C) 100 V (D) 10 V

Tardigrade Admin · Accepted Answer

F m = mg ∴ text ILB = mg ∴ (( V / R )) LB = mg ∴ V =( mgR / LB ) =((1 × 10-3 kg )(10 m / s 2)(10 Ω)/(0.1 m )(103 × 10-4 T ))=10 V

$\frac{1}{10} V$

$1 V$

$100 V$

$10 V$

$F_{m} = m g$
$∴ ILB = m g$
$∴ (\frac{V}{R}) L B = m g$
$∴ V = \frac{m g R}{L B}$
$= \frac{( 1 \times 1 0 ^{- 3} k g ) ( 10 m / s ^{2} ) ( 10Ω )}{( 0.1 m ) ( 1 0 ^{3} \times 1 0 ^{- 4} T )} = 10 V$

101​V

1V

100V

10V

Fm​=mg ∴ ILB =mg ∴(RV​)LB=mg ∴V=LBmgR​ =(0.1m)(103×10−4T)(1×10−3kg)(10m/s2)(10Ω)​=10V

$\frac{1}{10} V$

$1 V$

$100 V$

$10 V$

$F_{m} = m g$
$∴ ILB = m g$
$∴ (\frac{V}{R}) L B = m g$
$∴ V = \frac{m g R}{L B}$
$= \frac{( 1 \times 1 0 ^{- 3} k g ) ( 10 m / s ^{2} ) ( 10Ω )}{( 0.1 m ) ( 1 0 ^{3} \times 1 0 ^{- 4} T )} = 10 V$