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

Answer 1

Answer: (A)

0.5 T

Answer 2

Hence, m = 500g = $0.5$ kg, I= 4A, l =

As

$\left( \because \theta = 90 ^ { \circ } \right.$ and $\left. \mathrm { F } = \mathrm { mg } \right)$

$\therefore \mathrm { B } = \frac { \mathrm { mg } } { \mathrm { Il } } = \frac { 0.5 \times 10 } { 4 \times 2.5 } = 0.5 \mathrm {~T}$