Question

A straight wire (conductor) of length 10 cm is kept in a uniform magnetic field of induction 0.02 T. The angle between the conductor and the field direction is ${30}^0$ . A current of 5A is passed through the conductor. The force on the conductor is ( in Newton):

• A. $4\times {10}^{-3}$
• B. $5\times {10}^{-3}$
• C. $6\times {10}^{-3}$
• D. $7\times {10}^{-3}$

Answer 1

The correct option is B $5\times {10}^{-3}$

$\overline{F}=i(\overrightarrow{l}\times \overrightarrow{B})$
$F=BilSin\theta$
$=0\cdot 02\times 5\times 0\cdot 1\times sin{30}^0$
$=5\times {10}^{-3}N$