A 2.5 m long straight wire having mass of 500 g is suspended in mid air by a uniform horizontal magnetic field B. If a current of 4 A is passing through the wire then the magnitude of the field is (Take g=10 m s-2)

Question

A 2.5 m long straight wire having mass of 500 g is suspended in mid air by a uniform horizontal magnetic field B. If a current of 4 A is passing through the wire then the magnitude of the field is (Take g=10 m s-2) (A) 0.5 T (B) 0.6 T (C) 0.25 T (D) 0.8 T

Tardigrade Admin · Accepted Answer

Here, m=500 g =0.5 kg, I=4 A, l=2.5 m As F=IlBsinθ mg=IlB sin 90°, (∵ θ=90° and F=mg) ∴ B=(mg/Il)=(0.5×10/4×2.5)=0.5 T

Q. $A$ $2.5 m$ long straight wire having mass of $500 g$ is suspended in mid air by a uniform horizontal magnetic field $B$ . If a current of $4 A$ is passing through the wire then the magnitude of the field is (Take $g = 10 m s^{- 2})$

$0.5 T$

$0.6 T$

$0.25 T$

$0.8 T$

Here, $m = 500 g = 0.5 k g$ ,
$I = 4 A$ ,
$l = 2.5 m$
As $F = I lB s in θ$
$m g = I lB s in 9 0^{\circ}$ , $(∵ θ = 9 0^{\circ}$ and $F = m g)$
$∴ B = \frac{m g}{I l} = \frac{0.5 \times 10}{4 \times 2.5} = 0.5 T$

Q. A 2.5m long straight wire having mass of 500g is suspended in mid air by a uniform horizontal magnetic field B. If a current of 4A is passing through the wire then the magnitude of the field is (Take g=10ms−2)

0.5T

0.6T

0.25T

0.8T