For the adjoining figure, the magnetic field at a point 'P' will be:

1. $\frac{{\mu }_0}{4\pi }\odot$

2. $\frac{{\mu }_0}{\pi }\otimes$

3. $\frac{{\mu }_0}{2\pi }\otimes$

4. $\frac{{\mu }_0}{2\pi }\odot$

A charge having q/m equal to 10⁸ c/kg and with velocity 3 × 10⁵ m/s enters into a uniform magnetic field B = 0.3 tesla at an angle 30º with the direction of field. Then the radius of curvature will be:

1. 0.01 cm

2. 0.5 cm

3. 1 cm

4. 2 cm

An electron moves in a circular orbit with a uniform speed \(v\). It produces a magnetic field \(B\) at the centre of the circle. The radius of the circle is proportional to:
1. \(\sqrt{\frac{v}{B}}\)
2. \(\frac{v}{B}\)
3. \(\frac{B}{v}\)
4. \(\sqrt{\frac{B}{v}}\)

A very long straight wire carries a current I. At the instant when a charge +Q at point P has velocity $\overrightarrow{\mathrm{v}}$, as shown, the force on the charge is

1. Along ox

2. Opposite to oy

3. Along oy

4. Opposite to ox

Resistance of a Galvanometer coil is \(8~\Omega\) and \(2~\Omega\) shunt resistance is connected with it. If main current is \(1\) A then the current flow through \(2~\Omega\) resistance will be:
1. \(0.2\) A
2. \(0.8\) A
3. \(0.1\) A
4. \(0.4\) A

Two long parallel wires are at a distance of \(1\) m. If both of them carry one ampere of current in the same direction, then the force of attraction on the unit length of the wires will be:
1. \(2\times10^{-7}\) N/m
2. \(4\times10^{-7}\) N/m
3. \(8\times10^{-7}\) N/m
4. \(10^{-7}\) N/m

What properties will a galvanometer that is acting as a voltmeter have?

A current-carrying coil (I = 5A, R = 10 cm) has 50 turns. The magnetic field at its centre will be:
1.  1.57 mT
2.  3.14 mT
3.  1 mT
4.  2 mT

A galvanometer of \(50~\Omega\) resistance has \(25\) divisions. A current of \(4\times 10^{-4}~\text{A}\) gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of \(25~\text{V}\), it should be connected with a resistance of: