A rectangular coil of length 0.12 m and width 0.1 m having 50 turns of wire is suspended vertically in a uniform magnetic field of strength 0.2 Wb/m². The coil carries a current of 2 A. If the plane of the coil is inclined at an angle of 30^o with the direction of the field, the torque required to keep the coil in stable equilibrium will be:

1. 0.15 Nm

2. 0.20 Nm

3. 0.24 Nm

4. 0.12 Nm

Two identical long conducting wires AOB and COD are placed at the right angle to each other, with one above other such that 'O' is their common point for the two. The wires carry I₁ and I₂ currents, respectively. Point 'P' is lying at distance 'd' from 'O' along a direction perpendicular to the plane containing the wires. The magnetic field at the point 'P' will be :

1. $\frac{{\mu }_0}{2\mathrm{\pi d}}\left(\frac{I_1}{I_2}\right)$

2. $\frac{{\mu }_0}{2\mathrm{\pi d}}\left(I_1+I_2\right)$

3. $\frac{{\mu }_0}{2\mathrm{\pi d}}\left(I_1^2+I_2^2\right)$

4. $\frac{{\mu }_0}{2\mathrm{\pi d}}{\left(I_1^2+I_2^2\right)}^{1/2}$

Two toroids 1 and 2 have total no. of turns 200 and 100 respectively with average radii 40 cm and 20 cm respectively. If they carry the same current i, the ratio of the magnetic fields along the two loops is:

(1) 1:1

(2) 4:1

(3) 2:1

(4) 1:2

A galvanometer has a coil of resistance 100 and gives a full-scale deflection for 30 mA current. If it is to work as a voltmeter of 30 V range, the resistance required to be added will be:

1. 900 $\Omega$

2. 1800 $\Omega$

3. 500 $\Omega$

4. 1000 $\Omega$

A current-carrying closed loop in the form of a right isosceles triangle ABC is placed in a uniform magnetic field acting along AB. If the magnetic force on the arm BC is F, the force on the arm AC is :
                

1.  $-\mathrm{F}$

2.  $\mathrm{F}$

3. $2\mathrm{F}$

4.  $-2\mathrm{F}$

A charged particle (charge q) is moving in a circle of radius R with uniform speed v. The associated magnetic moment μ is given by:

1. $\frac{qvR}{2}$

2. $qv{R}^2$

3. $\frac{qv{R}^2}{2}$

4. $qvR$

A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in the equilibrium state. The energy required to rotate it by 60^o is W. Now the torque required to keep the magnet in this new position is:

1.  $\frac{W}{\sqrt{3}}$

2.  

3.  

4.  

A millivoltmeter of 25 mV range is to be converted into an ammeter of 25 A range. The value (in ohm) of necessary shunt will be:

1. 0.001

2. 0.01

3. 1

4. 0.05