CBSE
Gujarat Board
Haryana Board
Class 10
Class 12
A closely wound solenoid of 2000 turns and area of cross -sectionÂ 1.5 x 10^{-4}Â m^{2} carries a current of 2.0 A. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field 5 x 10^{-2} T making an angle of 30^{o} with axis of Â the solenoid. The torque on the solenoid will be
3 x 10^{-3} N-m
1.5 x 10^{-3} N-m
1.5 x 10^{-2} N-m
3 x 10^{-2} N-m
A.
3 x 10^{-3} N-m
Given, N = 2000,
A = 1.5 x 10^{-4} m^{2}^{}
i = 2.0 A B = 5 x 10^{-2} T
and
Î¸ = 30^{o}
Torque = NiBA sinÂ Î¸
2000 x 2 x 5 x 10^{-2} x 1.5 x 10^{-14} x sin 30^{o}
= 2000 x 50 x 10^{-6} x (1/2)
= 1.5 x 10^{-2} Nm
Given, N = 2000,
A = 1.5 x 10^{-4} m^{2}^{}
i = 2.0 A B = 5 x 10^{-2} T
and
Î¸ = 30^{o}
Torque = NiBA sinÂ Î¸
2000 x 2 x 5 x 10^{-2} x 1.5 x 10^{-14} x sin 30^{o}
= 2000 x 50 x 10^{-6} x (1/2)
= 1.5 x 10^{-2} Nm