In a cyclotron, the angular frequency of a charged particle is independent of :

1. Mass                         

2.  Speed

3. Charge                       

4. Magnetic field

Two particles A and B of masses $m_A$ and $m_B$ respectively and having the same type of charge are moving in a plane. A uniform magnetic field exists perpendicular to this plane. The speeds of the particles are $v_A$ and $v_B$ respectively, and the trajectories are as shown in the figure. Then

1. $m_A{v}_A<m_B{v}_B$                              

2. $m_A{v}_A>m_B{v}_B$

3. $m_A<m_B and v_A<v_B$                  

4. $m_A=m_B and v_A=v_B$

In a moving coil galvanometer, the deflection of the coil $\theta$ is related to the electrical current i by the relation

1. $i\propto \tan \theta$                 

2. $i\propto \theta$ 

3. $i\propto {\theta }^2$                    

4. $i\propto \sqrt{\theta }$ 

A particle with charge \(q\), moving with a momentum \(p\), enters a uniform magnetic field normally. The magnetic field has magnitude \(B\) and is confined to a region of width \(d\), where \(d< \frac{p}{Bq}.\) The particle is deflected by an angle \(\theta\) in crossing the field, then:

A current \(I\) is carried by an elastic circular wire of length \(L\). It is placed in a uniform magnetic field \(B\) (out of paper) with its plane perpendicular to \(B'\text{s}\) direction. What will happen to the wire?

Three long, straight parallel wires carrying current, are arranged as shown in figure. The force experienced by a 25 cm length of wire C is

1. ${10}^{-3}N$

2. $2.5\times {10}^{-3}N$

3. Zero

4. $1.5\times {10}^{-3}N$

A wire carrying current l has the shape as shown in the adjoining figure. Linear parts of the wire are very long and parallel to X-axis while the semicircular portion of radius R is lying in the Y-Z plane. Magnetic field at point O is :


1. $\mathrm{B}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\times \frac{\mathrm{i}}{\mathrm{R}}\left(\mathrm{\pi }\hat{\mathrm{i}}+2\hat{\mathrm{k}}\right)$

2. $\mathrm{B}=-\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\times \frac{\mathrm{i}}{\mathrm{R}}\left(\mathrm{\pi }\hat{\mathrm{i}}-2\hat{\mathrm{k}}\right)$

3. $\mathrm{B}=-\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\times \frac{\mathrm{i}}{\mathrm{R}}\left(\mathrm{\pi }\hat{\mathrm{i}}+2\hat{\mathrm{k}}\right)$

4. $\mathrm{B}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\times \frac{\mathrm{i}}{\mathrm{R}}\left(\mathrm{\pi }\hat{\mathrm{i}}-2\hat{\mathrm{k}}\right)$

A part of a long wire carrying a current i is bent into a circle of radius r as shown in the figure. The net magnetic field at the centre O of the circular loop is

1. $\frac{{\mu }_{0}i}{4r}$                                

2. $\frac{{\mu }_{0}i}{2\mathrm{\pi r}}\left(\mathrm{\pi }+1\right)$

3. $\frac{{\mu }_{0}i}{2r}$                                

4. $\frac{{\mu }_{0}i}{2\mathrm{\pi r}}\left(\mathrm{\pi }-1\right)$