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

Magnet Quiz 2

physics

 The magnetic flux through any surface will always be zero. While considering this statement you will come across a law. This law is given by which of the following equations?

 Think of the law with respect to the magnetism possessing $m$ as the surface integral of the magnetic field over any surface.

  The law that we are considering in the given question is basically: ‘The Gauss’s law in magnetism’. $ \\ $ This law states that the net magnetic flux through any surface whether open or closed is always zero. It is given by the equation: $ \\ \oint_{S}{B.dS=0} \\ $ Also, the surface integral $\oint{_{S}}$ of the magnetic field $B$ is always $m$ over any closed or open surface $dS$.

 While making a magnetic map of earth, we draw some lines on the earth’s surface. Of those lines, one line is such that it joins places of zero inclination. What is the name of such a line?

 Recall the various types of the magnetic lines around the earth’s surface that forms the magnetic map. Of those lines you will find a line that will join places of dip on the earth’s surface.

  When we talk of magnetic maps, we come across various lines that are found on the surface of the earth. These lines pass through different places and possess the same magnetic elements. If we consider a line that joins places of zero inclination or dip, such a line would be called the ‘Aclinic line’.

 What will be the magnetisation due to a coil with $500\text{ turns}/m$ is carrying a current of $1.2\,A$ ? $($Permeability of the material from which the coil was made is $200)$

 Magnetisation of a material is equal to its net magnetic moment per unit volume. If solenoid of $n$ turns per unit length and carries a current $I$. The magnetic field in the interior of the solenoid was shown to be given by, $M=\dfrac{\left( B-{{\mu }_{0}}nI \right)}{{{\mu }_{0}}}$.

 Magnetisation of a coil is given by,  $ \\ M=\dfrac{\left( B-{{\mu }_{0}}H \right)}{{{\mu }_{0}}} \\ $ $B$ is given by the formula, $ \\ B={{\mu }_{r}}{{\mu }_{0}}H   --(1) \\ $ $H$ is independent of the type of material from which the coil was made and it is given by, $ \\ H=nI  --(2) \\ $  Now using $(1)$ and $(2)$, Magnetisation of a coil will be, $ \\ M=\dfrac{\left( {{\mu }_{r}}{{\mu }_{0}}H-{{\mu }_{0}}H \right)}{{{\mu }_{0}}} \\ \Rightarrow M=\dfrac{{{\mu }_{0}}H\left( {{\mu }_{r}}-1 \right)}{{{\mu }_{0}}}=H\left( {{\mu }_{r}}-1 \right) \\ \Rightarrow M=\left( {{\mu }_{r}}-1 \right)nI \\ \therefore M=\left( 200-1 \right)\times 500\times 1.2=1.19\times {{10}^{5}}\,A/m$

  In an experiment to check the points without a magnetic field by changing the direction of the magnet, a student placed a magnet along the magnetic north-south direction. A point where no magnetic field is observed on the horizontal axis at 18 cm from the centre of the magnet. Then, what will be the magnetic field on the vertical axis of the magnet at 18 cm from the centre of the magnet? $($At the laboratory, Earth’s magnetic field is $0.48\text{ }G$ and the angle of dip is zero$)$

 On null points along the horizontal axis, earth’s field and magnet’s field will be opposite to each other and on the vertical axis, magnet’s field will be opposite to its field on the axis. So, they get added up.

 According to the question, the point at $18\,cm$ along the horizontal axis is a null point. Null point along the axis is given by the equation, $ \\ {{B}_{1}}=\dfrac{{{\mu }_{0}}}{4\pi }\dfrac{2M}{{{d}^{3}}}=H=0.48\,G \\ $  On the vertical axis of the magnet at $18cm$, the field due to the magnet will be half of the field on the horizontal axis at the same distance. $ \\ {{B}_{2}}=\dfrac{{{\mu }_{0}}}{4\pi }\dfrac{M}{{{d}^{3}}}=\dfrac{H}{2} \\ $ This field will be opposite to ${{B}_{1}}$. So they get added up. $ \\ \Rightarrow B={{B}_{1}}+{{B}_{2}} \\ \Rightarrow B=H+\dfrac{H}{2}=\dfrac{3H}{2} \\ \therefore B=\dfrac{3\times 0.48}{2}=0.72\,G$

  In Tamil Nadu, assume that earth’s magnetic field $0.26\,G$, the angle of dip is $28{}^\circ $ and magnetic declination is zero. $4$ high-voltage cable (HV cable) carrying a current of $10.0\,A$ from Chennai to Coimbatore.  What will be the magnetic field $1\,m$ away from these cables?

 Find the magnitude of horizontal and vertical components of earth’s magnetic field and magnetic field due to the cables using formula from biot-savart’s law. Now, using vector addition after understanding right directions, find the resultant field.

 It is given that the cable carries a total current of $40.0\,A$ in the direction east to west $($Chennai to Coimbatore$)$. So the resultant magnetic field 1 m away from the cables will be, $ \\ {{B}_{net}}=\sqrt{{{\left( {{B}_{H}}-{{B}_{wire}} \right)}^{2}}+{{B}_{V}}^{2}} \\ $ Here, horizontal component of earth’s magnetic field is, $ \\ {{B}_{H}}=B\cos \delta  \\ \therefore {{B}_{H}}=0.26\times \cos 28{}^\circ =0.23\,G \\ $ Here, vertical component of earth’s magnetic field is, $ \\ {{B}_{V}}=Bsin\delta  \\ \therefore {{B}_{V}}=0.26\times \sin 28{}^\circ =0.12\,G \\ $ Now, magnetic field due to the wire will be, $ \\ {{B}_{wire}}=\dfrac{{{\mu }_{0}}}{4\pi }\dfrac{2I}{r} \\ \Rightarrow {{B}_{wire}}={{10}^{-7}}\times \dfrac{2\times 40}{1} \\ \therefore {{B}_{wire}}=0.8\times {{10}^{-5}}T=0.08\,G \\ $ Now, resultant field will be, $ \\ {{B}_{net}}=\sqrt{{{\left( 0.23-0.08 \right)}^{2}}+{{\left( 0.12 \right)}^{2}}} \\ \Rightarrow {{B}_{net}}=\sqrt{{{\left( 0.222 \right)}^{2}}+{{\left( 0.12 \right)}^{2}}} \\ \therefore {{B}_{net}}=0.252\,G$