Question

A torque of 25 N m acts on a current carrying coil of area ${5m}^2$ in a magnetic field of induction ${2Wb/m}^2$ . The angle between normal to coil and magnetic induction is ${30}^{\circ }$ .Then value of current is

• A. $0.4A$
• B. $0.5A$
• C. $400mA$
• D. $5A$

Answer 1

The correct option is D $5A$

Given: $\tau =25Nm$

$A=5m^2$ \\area of coil

$B=2Wb/m^2$

$\theta ={30}^{\circ }$

To find: current,$I=?$

Solution: $\tau =IABsin\theta$

$==>25=I\ast 5\ast 2\ast sin{30}^{\circ }$

$==>5=I\ast 2\ast \left(1/2\right)$

$I=5Amp$

hence,

The correct opt : D