A circular coil of \(30\) turns and a radius of \(8.0 ~\text{cm}\) carrying a current of \(6.0 ~\text{A}\) is suspended vertically in a uniform horizontal magnetic field of magnitude \(1.0 ~\text{T}.\) The field lines make an angle of \(60^\circ\) with the normal of the coil. What will be the magnitude of the counter-torque that must be applied to prevent the coil from turning?
1. \(7.12 ~\text{N-m}\)
2. \(3.13~\text{N-m}\)
3. \(6.50~\text{N-m}\)
4. \(4.44~\text{N-m}\)

A square coil of side \(10~\text{cm}\) consists of \(20~\text{turns}\) and carries a current of \(12~\text{A}.\) The coil is suspended vertically and the normal to the plane of the coil makes an angle of \(30^{\circ}\) with the direction of a uniform horizontal magnetic field of magnitude \(0.80~\text{T}.\) What is the magnitude of torque experienced by the coil?
1. \(0.79~\text{N-m}\)
2. \(0.88~\text{N-m}\)
3. \(0.49~\text{N-m}\)
4. \(0.96~\text{N-m}\)

A circular coil of \(20\) turns and a radius of \(10~\text{cm}\) is placed in a uniform magnetic field of \(0.10~\text{T}\) normal to the plane of the coil. If the current in the coil is \(5.0~\text{A},\) what is the total torque on the coil?
1. \(1.0~\text{N-m}\)
2. \(\text{zero}\)
3. \(0.5~\text{N-m}\)
4. \(0.3~\text{N-m}\)

A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is \(60^{\circ}\), and one of the fields has a magnitude of \(1.2\times 10^{-2}~\text{T}\). If the dipole comes to stable equilibrium at an angle of \(15^{\circ}\) with this field, what is the magnitude of the other field?  \(\left[\text{Given} :   \sin   15^ \circ = 0 . 26\right]\)
1. \( 7.29 \times10^{-3} ~\text{T} \)
2. \( 4.39 \times10^{-3} ~\text{T} \)
3. \( 6.18 \times10^{-3} ~\text{T} \)
4. \(5.37 \times10^{-3} ~\text{T} \)