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This is second of our many quizzes on Magnetism. Click on this to access more such competitive quizzes on Electrostatics and Magnetism.

Magnetism Quiz 2

physics

An electron enters in a uniform magnetic field directed outward into the plane. If the velocity of electron is $V$ then direction of force act on it is in $ \\ $ <br><img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/4f34e43a-732d-439d-8c44-0c5aa48adcb82255535287726269484.png"/>

Use the formula $\overrightarrow{f}=q\left( \overrightarrow{V}\times \overrightarrow{B} \right)$ and put the value of $''q''$ sign wise. 

  $\text{For electron charge is negative} \\ \overrightarrow{F}=-q\left( \overrightarrow{V}\times \overrightarrow{B} \right) \\ \Rightarrow\overrightarrow{F}=-q\left( {{V}_{0}}\hat{j}\times {{B}_{0}}\hat{k} \right) \\ \Rightarrow\overrightarrow{F}=-q\left( {{V}_{0}}{{B}_{0}} \right)\left( \hat{j}\times \hat{k} \right) \\ \therefore\overrightarrow{F}=-q{{V}_{0}}{{B}_{0}}\hat{i} \\ \text{Thus, the force is in }-\hat{i}\text{ or negative }x-axis\text{ direction}\text{.} $

A magnetic dipole is placed perpendicular to a magnetic field and then immediately released. If the dipole moment of the dipole is $\overrightarrow{m}$, and the magnetic field is $\overrightarrow{B},$ then the angular velocity of the dipole when it is parallel to the magnetic field. The dipole is in the shape of a rod of length $\ell $ and mass $M$. 

As the magnetic field is conservative, total mechanical energy is conserved. Potential energy of a dipole $=-\overrightarrow{m}.\overrightarrow{B}$ and kinetic energy of a rod $=\dfrac{1}{2}I{{\omega }^{2}}$, $I$ is moment of Inertia, $\omega $ is angular frequency.

  $\text{Initially }\theta =90{}^\circ \text{ between }\overrightarrow{m}\text{ and }\overrightarrow{B}. \\{{U}_{i}}=-\overrightarrow{m}.\overrightarrow{B}=-mB\cos 90{}^\circ =0 \\ \text{Initial kinetic energy} \\{{K}_{i}}=0 \\ \text{When }\theta =0{}^\circ \text{ between }\overrightarrow{m}\text{ and }\overrightarrow{B}\text{ after rotation,} \\{{U}_{f}}=-\overrightarrow{m}.\overrightarrow{B}=-mB\cos 0{}^\circ =-mB \\ \Rightarrow{{K}_{f}}=\dfrac{1}{2}I{{\omega }^{2}} \\ \text{ is moment of inertia about its centre}\text{.} \\I\text{ for a rod about its centre}=\dfrac{1}{2}M{{\ell }^{2}} \\ \text{As magnetic field is conservative in nature,} \\{{U}_{i}}+{{K}_{i}}={{U}_{f}}+{{K}_{f}} \\ \Rightarrow 0+0=-mB+\dfrac{1}{2}I{{\omega }^{2}} \\ \Rightarrow \dfrac{1}{2}I{{\omega }^{2}}=mB \\ \Rightarrow {{\omega }^{2}}=\dfrac{2mB}{I}=\dfrac{2mb}{\dfrac{1}{12}m{{\ell }^{2}}} \\ \therefore \omega =\dfrac{2}{\ell }\sqrt{\dfrac{6mB}{M}}$

An imaginary sphere of radius $''r''$ is imagined with its centre at the edge $''O''$ of a solenoid. If the solenoid has $n$ turns per unit length and current $I$ flows through it, then what is the total magnetic flux through the sphere is<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/2750a5d5-633f-40e9-b800-d6a04dd159777001983053731783379.png"/> 

 Use gauss law of magnetism, which deals with the magnetic flux, area and magnetic field as $\phi =\overrightarrow{B}.\overrightarrow{A}.$ 

 For a closed surface, gauss law of magnetic field, says that the total magnetic flux is always zero, as the magnetic field lines enter and exit in the same number. In other words. $ \\{{\phi }_{in}}=-{{\phi }_{out}} \\ \Rightarrow {{\phi }_{total}}={{\phi }_{in}}+{{\phi }_{out}} \\ \therefore {{\phi }_{total}}=0$

 If permeability of an alloy increases then $x$ susceptibility ______. 

Magnetic susceptibility varies as permeability ${{\mu }_{r}}=1+\chi $ i.e. ${{\mu }_{r}}\propto \chi $ use this relation to get the answer.  

 ${{\mu }_{r}}=1+\chi  \\ \therefore{{\mu }_{r}}\propto \chi \\ $ It means when permeability is increase, then susceptibility will also increases.

A bar magnet made up of conducting material freely falls in the line of axis perpendicular to the centre of a metal ring. The acceleration position graph for the motion of the bar magnet is- $(\text{If it’s position is measured vertically})$ 

The change in flux is more when the magnet is nearer to the ring. This causes greater induced current. Thus, an opposing force against the motion of the magnet comes up, which is stronger the nearer the magnet. 

 Initially, the freely falling magnet will have a small opposing force as it is falling down and inducing a current in the metal ring.As the magnet approaches the ring the opposing force grows stronger, thus the total acceleration of the magnet decreases more steeply.Once the magnet passes through the ring, the opposing force will change direction and still be opposite the gravity. The curve will be symmetric, and the acceleration again becomes $g$.