AP EAMCET202022 Sep 2020Evening ShiftPhysicsMagnetic Effects of CurrentActual
A bar magnet of length 10 ~cm and having the pole strength equal to 10⁻³ ~A - m is kept in a magnetic field having magnetic induction B equal to 4 10⁻³ ~T . It makes an angle of 30^ with the direction of magnetic induction. The value of the torque acting on the magnet is
Options
- A2 10⁻⁷ Nm
- B2 10⁻⁵ Nm
- C0.5 Nm
- D0.5 10^2 Nm
Correct answer
A. 2 10⁻⁷ Nm
Step-by-step solution
Given, length of bar magnet, l=10 ~cm =10⁻¹ ~m Pole strength, m=10⁻³ ~A - m aligned & B=4 10⁻³ ~T & =30^ aligned Magnetic dipole moment, M=m l=10⁻³ 10⁻¹=10⁻⁴ ~A - m ^2 Torque, =M B aligned & =10⁻⁴ 4 10⁻³ 30^ & =4 10⁻⁷ 1 2 =2 10⁻⁷ ~N - m aligned