A straight wire of length 50 cm carrying a current of 2.5 A is suspended in mid-air by a uniform magnetic field of 0.5 T (as shown in figure). The mass of the wire is (g = 10 ms-2 )

Question

A straight wire of length 50 cm carrying a current of 2.5 A is suspended in mid-air by a uniform magnetic field of 0.5 T (as shown in figure). The mass of the wire is (g = 10 ms-2 ) (A) 62.5 gm (B) 250 gm (C) 125 gm (D) 100 gm

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/32557d4dc6e672e1b9d8838b80e52dbf-.png" /> Given, length of wire (I)=50 cm =50 × 10-2 m current (I)=2.5 A magnetic field (B)=0.5 T g=10 ms -2 For the balance condition, FB =mg BIl =m g m =(B I l/g) =(0.5 × 2.5 × 50 × 10-2/10) =62.5 g

Q. A straight wire of length $50 c m$ carrying a current of $2.5 A$ is suspended in mid-air by a uniform magnetic field of $0.5 T$ (as shown in figure). The mass of the wire is $(g = 10 m s^{- 2})$

62.5 gm

250 gm

125 gm

100 gm

Given, length of wire $(I) = 50 c m = 50 \times 1 0^{- 2} m$
current $(I) = 2.5 A$
magnetic field $(B) = 0.5 T$
$g = 10 m s^{- 2}$
For the balance condition,
$F_{B} = m g$
$B I l = m g$
$m = \frac{B I l}{g}$
$= \frac{0.5 \times 2.5 \times 50 \times 1 0 ^{- 2}}{10}$
$= 62.5 g$

Q. A straight wire of length 50cm carrying a current of 2.5A is suspended in mid-air by a uniform magnetic field of 0.5T (as shown in figure). The mass of the wire is (g=10ms−2)