MAGNETISM

Magnetic field at the center:

          The distance between current element on the circular coil and the center is ‘r’ and the angle between current element and ‘r’ is 90^o

By applying Biot Savart’s Law –

                                                \(dB=\frac{{{\mu }^{0}}}{4\pi }\frac{idl\sin \theta }{{{r}^{2}}}\)

Here,

θ = 90^o

                                                                                          ^{\(\therefore dB=\frac{{{\mu }^{0}}}{4\pi }\frac{idl\sin 90}{{{r}^{2}}}\)}

                                                          \(\therefore dB=\frac{{{\mu }^{0}}}{4\pi }\frac{idl}{{{r}^{2}}}\)

The complete circular coil is a combination of multiple current elements. So to find the resultant magnetic field due to all the current elements at the center we need to integrate dB.

                                                \(\begin{align} & \int{dB=\oint ~}\frac{{{\mu }^{0}}}{4\pi }\frac{idl}{{{r}^{2}}} \\ & B=\frac{{{\mu }^{0}}}{4\pi }\frac{i}{{{r}^{2}}}2\pi r \\ & B=\frac{{{\mu }^{0}}i}{2r} \\ \end{align}\)

Direction of magnetic field at the centre is outwards.         

By observing magnetic field along the complete axis we conclude that any charge moving in the circle behaves as a magnet.

Coil behaves like a bar magnet.

1.         A small hole is made at the centre of the magnet then its magnetic

Solution:

                        Decreases

2.         The dimensional formula for magnetic moment is

Solution:

                        M0L2T0A1  

1.         A charge q moves in a circular path of radius r with a speed of v. Calculate the induction of the magnetic field produced at                   the centre of the circle.

Solution:

            The equivalent current in a circular path is

                                    i = \(\frac{q}{t}=\frac{q}{2\pi (r/v)}=\frac{qv}{2\pi r}\)
            Hence the induction of the magnetic field at the centre of the circle is
                                    B = \(\left( \frac{{{\mu }_{0}}}{4\pi } \right)\frac{2\pi i}{r}=\left( \frac{{{\mu }_{0}}}{4\pi } \right)\frac{2\pi qv}{2\pi {{r}^{2}}}=\left( \frac{{{\mu }_{0}}}{4\pi } \right)\frac{qv}{{{r}^{2}}}\) Tesla, along the axis.

   2.         A proton of velocity 1.0 × 107 m/s is projected at right angles to a uniform magnetic field of induction 100 wb/m2.

               How long does it take for the proton to traverse a 90o arc? (mp = 1.65 × 10-27 kg, q = 1.6 × 10-19 C).

Solution:

                Let 't' be the time taken by the particle to traverse a 90° arc.  Then

                                              \(t=\frac{\pi }{2}\,.\,\frac{m}{qB}\,\,=\,\,1.619\,\times \,{{10}^{-10}}s\)

                                                            \(\,as\,\,\,B=\frac{{{\mu }_{o}}I\theta }{4\pi r}=0\)

1.          A proton of velocity 1.0 × 107 m/s is projected at right angles to a uniform magnetic field of induction 100 wb/m2. How much               is the particle path deflected from a straight line after it has traversed a distance of 1 cm in the direction of initial velocity.

Solution:

             It is clear that the proton would describe a circular path under the given conditions.

             The radius of the path is given by     

                           \(x=\frac{1\,\,\times \,\,{{10}^{-4}}}{2r}\,=\,\frac{1\,\,\times \,{{10}^{-4}}}{2\,\,\times \,1.6576}\)  =  4.848 × 10-5 m.

2.         A proton of velocity 1.0 × 107 m/s is projected at right angles to a uniform magnetic field of induction 100 wb/m2. How long                  does it take for the proton to traverse a 90o arc? (mp = 1.65 × 10-27 kg, q = 1.6 × 10-19 C).

Solution:

               Let 't' be the time taken by the particle to traverse a 90° arc.  Then

                                              \(t=\frac{\pi }{2}\,.\,\frac{m}{qB}\,\,=\,\,1.619\,\times \,{{10}^{-10}}s\)

1.       Shown in the figure is a conductor carrying a current i. Find the magnetic field intensity at the point O.

Solution:

                               Net magnetic field  \(\vec{B}\) = \({{\vec{B}}_{1}}\)+ \({{\vec{B}}_{2}}\,\,+\,\,{{\vec{B}}_{3}}\)

                               Hence net \(\vec{B}\) = \(\frac{{{\mu }_{o}}I}{4\pi }\,\,\left[ -\frac{1}{r}+\frac{1}{2r}-\frac{1}{3r} \right]\,\theta =-\frac{5{{\mu }_{o}}I\theta }{24\pi r}\,\hat{k}\)  

2.      Current I flows through the wire as shown in the figure. The semi-circular part of the wire is perpendicular to the x-y plane.                   Find the magnetic field at the centre C of the semi-circle.

Solution:

              First we determine the direction of the magnetic field due to the straight parts of the wire. Magnetic field due to the two straight parts is directed into the page i.e. along negative z-axis. Magnetic field due to the semi-circular part is directed along CO. Now magnitude of the magnetic field due to any one of the straight parts is

                                                \({{B}_{1}}\ =\ \frac{{{\mu }_{0}}\ I}{4\pi \ a}\ \left[ \operatorname{Sin}\ {{90}^{o}}-\operatorname{Sin}\ {{45}^{o}} \right]\,\,\,\)

                                                         = \(\frac{{{\mu }_{0}}\ I}{4\pi \ a}\ \left( 1-1/\sqrt{2} \right)\)                     (along z-direction)

                                Magnetic field due to the semi-circular part BZ  = \(\frac{{{\mu }_{0}}\ I}{2\ \sqrt{2}\ a}\) (Radius of the semi circle is \(\sqrt{2}\ a\))

                                Now unit vector along CO = ‑ \(\frac{1}{\sqrt{2}}\ (\hat{i}+\hat{j})\)

                                \           The resultant magnetic field at C.

                                                 \(\vec{B}\ =\ 2\frac{{{\mu }_{0}}\ I}{4\pi \ a}\left( \frac{1}{\sqrt{2}}-1 \right)\ \hat{k}-\frac{{{\mu }_{0}}\ I}{4\ a}(\hat{i}+\hat{j})\)

                                               = ‑ \(\frac{{{\mu }_{0}}\ I}{4\ a}\left[ \hat{i}+\hat{j}+\frac{(2-\sqrt{2})}{\pi }\hat{k} \right]\).

MAGNETISM

Suppose a straight current carrying wire AB, carrying current i, lies in the plane of the paper as shown in the figure, P is a point at a perpendicular distance R from conductor, where magnetic field is to be determined.

According to Biot-Savart’s Law, the field at point P due to current element idl = qVB→ is

                                                       \(dB=\frac{{{\mu }_{0}}}{4\pi }\frac{idl\sin \theta }{{{r}^{2}}}=\frac{{{\mu }_{0}}I}{4\pi }(\frac{dl\sin \theta }{r})\frac{i}{r}\)

Now from the figure dlsinθ = rdφ and cosφ = R/r

                                                    \(\therefore dB=\frac{{{\mu }_{0}}I}{4\pi }.\frac{\cos \varphi }{R}d\varphi \)

Hence magnetic field due to the whole conductor

                        B = θ₁∫−θ₂ = dB = θ₁∫−θ₂ \(\frac{{{\mu }_{0}}I}{4\pi R}cos\varphi .d\varphi \text{ }=~\frac{{{\mu }_{0}}I}{4\pi R}[sin{{\theta }_{1}}~+\text{ }sin{{\theta }_{2}}]\)

                                                                                 _{\(\therefore dB=~\frac{{{\mu }_{0}}I}{4\pi R}[sin{{\theta }_{1}}~+\text{ }sin{{\theta }_{2}}]\)}

Straight current carrying conductor of infinite length –

                                                            θ₁ = θ₂ = π/2

                                                            \( \therefore B=\frac{{{\mu }_{0}}I}{2\pi R}\)

Magnetic field on perpendicular bisector of a conductor of finite length –

                                                \(\begin{align} & \sin {{\theta }_{1}}=\sin {{\theta }_{2}}=\frac{a/2}{\surd {{\left( a/2 \right)}^{2}}+{{d}^{2}}}=\frac{a}{\surd {{a}^{2}}+4{{d}^{2}}} \\ & \therefore \text{ }Magnetic\text{ }Field\text{ }B\text{ }=~\frac{{{\mu }_{0}}I}{4\pi R}\frac{2a}{\surd {{a}^{2}}+4{{d}^{2}}} \\ \end{align}\)

Where,

a = Length of the wire.

d = Perpendicular distance of the point P.

Magnetic field at point exactly in front of the end of a semi-infinite wire –

                                                     \(\begin{align} & {{\mathbf{\theta }}_{\mathbf{1}}}~=\text{ }\mathbf{0}~and~{{\mathbf{\theta }}_{\mathbf{2}}}~=\text{ }\mathbf{\pi }/\mathbf{2} \\ & \therefore \text{ }B\text{ }=\frac{{{\mu }_{0}}I}{4\pi R}\left( sin0\text{ }+\text{ }sin\pi /2 \right)\text{ }=\frac{{{\mu }_{0}}I}{4\pi R} \\ \end{align}\)

Magnetic field at a point not exactly in front of the end of a semi-infinite wire –

                                                       \(\begin{align} & {{\mathbf{\theta }}_{\mathbf{1}}}~=\text{ }\mathbf{\alpha }~and~{{\mathbf{\theta }}_{\mathbf{2}}}~=\text{ }\mathbf{\pi }/\mathbf{2} \\ & B\text{ }=\frac{{{\mu }_{0}}I}{4\pi R}\left( sin\alpha \text{ }+\text{ }sin\pi /2 \right) \\ & \therefore B\text{ }=\frac{{{\mu }_{0}}I}{4\pi R}\left( 1\text{ }+\text{ }sin\alpha \right) \\ \end{align}\)

1.           The total number of magnetic lines of force originating or terminating on a pole of

              strength ‘m’ is

Solution:

                  \({{\mu }_{0}}m\)

2.         The electric  and magnetic field lines differ in that

Solution:

                      Magnetic field lines are closed while electric lines are not

1.    A straight conductor of mass m and carrying a current i is hinged at one end and placed in a plane perpendicular to the                       magnetic field of intensity \(\vec{B}\) as shown in the figure. At any moment if the conductor is let free, then the angular acceleration of           the conductor will be :

Solution:

                     The force acting on the elementary portion of the current carrying conductor is given as,

                              DF = i(dl)B sin 900 Þ dF = iBdr

                              The torque applied by dF about O = dt = rdF

                           Þ  The total torque about O = t

                                  \(\int{d\tau =\int{r(iBdr)}}\)

                                      \(\Rightarrow \,\,\,\tau =iB\int\limits_{0}^{L}{rdr}=\frac{iB{{L}^{2}}}{2}\)

                          The angular acceleration \(\frac{\tau }{M.I.}\)

                                        \(\Rightarrow \,\,\alpha ={\left( \frac{iB{{L}^{2}}}{2} \right)}/{\left( \frac{m{{L}^{2}}}{3} \right)}\;\)

                                        \(\Rightarrow \alpha =\frac{3iB}{2m}\)

2.    Find the magnetic field on the aaxis of a solenoid having n turns per unit length and carryng a current I.

Solution :     

                                        dB = \(\frac{{{\mu }_{0}}\,NI{{R}^{2}}}{2\,{{\left( {{R}^{2}}\,+\,{{x}^{2}} \right)}^{{\,3}/{2}\;}}}\)

                                        Here, N = ndx, x = R tanf  and dx = R sec2f df

                                        Hence,  dB = \(\frac{{{\mu }_{0}}\,ndx\,\,x\,\,I{{R}^{2}}}{2\,{{\left( {{R}^{2}}\,+\,{{R}^{2}}{{\tan }^{2}}\varphi \right)}^{{3}/{2}\;}}}\)

                                        \(B\,\,=\,\,\int{\,dB\,\,=\,\,\frac{{{\mu }_{0}}}{2}\,nI\,\int_{-\alpha }^{\beta }{\cos \,\varphi \,d\varphi }}\)

                                        i.e. B = \(\frac{{{\mu }_{0}}\,n\,I}{2}\,\left[ \sin \,\alpha \,+\,\sin \,\beta \right]\)

1.         Shown in the figure is a conductor carrying a current i. The magnitude of magnetic field at the origin is:

2.      Shown in the figure is a very long semicylindrical conducting shell of radius R and carrying a current i. An infinitely long                     straight current carrying conductor is lying   along the axis of the semicylinder. If the current flowing through the straight wire            be io, then the force on the semicylinder is:

Solution :     

1.      A wire loop carrying a current I is placed in the x-y plane as shown in fig.

       a) If a particle with charge q and mass m is placed at the centre P and given a  velocity v along NP find its instantaneous acceleration.

             (b) If an external uniform magnetic induction \(\vec{B}\,=B\hat{i}\) is applied, find the force and torque acting on the loop.

Solution:

             (a) As in case of current-carrying straight conductor and arc, the magnitude of B is given by

                     B₁ = \(\frac{{{\mu }_{0}}\,I}{4\pi \,d}\,\left( \sin \,\alpha \,\,+\,\,\sin \beta \right)\)

And     B2 = \(\frac{{{\mu }_{0}}\,I\,\varphi }{4\,\pi \,r}\)

So in accordance with Right hang screw roule,

            \(\left( \overrightarrow{B}{{\,}_{W}} \right)\,\,=\,\,\frac{{{\mu }_{0}}}{4\pi }\,\frac{1}{\left( a\,\cos \,60 \right)}\,\,x\,\,2\,\sin \,60\,(-\hat{k})\)

and     \({{\left( \overrightarrow{B}\, \right)}_{MN}}\,\,=\,\,\frac{{{\mu }_{0}}}{4\pi }\,\frac{I}{a}\,\,x\,\left( \frac{2}{3}\pi \right)\,(-\hat{k})\)

and  hence net  \(\overrightarrow{B}\) at P due to the given loop

            \(\vec{B}={{\vec{B}}_{w}}+{{\vec{B}}_{A}}\)     Þ     \(\overrightarrow{\,B\,}\,\,=\,\,\frac{{{\mu }_{0}}}{4\pi \,}\,\frac{2I}{a}\left[ \sqrt{3}\,\,-\,\,\frac{\pi }{3} \right]\,\left( -\hat{k} \right)\)        (i)

Now as force on charged particle in a magnetic fields is given by

            \(\overrightarrow{\,F\,}\,\,=\,\,q\left( \overrightarrow{v}x\overrightarrow{B} \right)\)

so here,     \(\overrightarrow{\,F\,}\,\,=\,\,q\,v\,B\,\sin \,{{90}^{0}}\,\,along\,\,PF\)

i.e.       \(\overrightarrow{F}\,\,=\,\,\frac{{{\mu }_{0}}}{4\pi }\,\frac{2\,qvI}{a}\,\left[ \sqrt{3}\,-\,\frac{\pi }{3} \right]\,\,\,along\,\,PF\)

and so   \(\vec{a}\,\,=\,\,\frac{\overrightarrow{F}}{m}\,\,=\,\,{{10}^{-7}}\,\frac{2\,qvI}{a}\,\left[ \sqrt{3}\,-\,\frac{\pi }{3} \right]\,\,\,along\,\,PF\)

(b)      As d\(\overrightarrow{\,F\,}\,\,=\,\,Id\,\overrightarrow{L\,}\,x\,\overrightarrow{\,B\,},\,\,\,so\,\,\,\overrightarrow{\,F\,}\,\,=\,\,\int_{{}}^{{}}{Id\overrightarrow{\,L\,}\,x\,\overrightarrow{\,B\,}}\)

          As here I and \(\overrightarrow{B}\) are constant

                     \(\overrightarrow{F}\,=\,\,I\,\left[ \oint{d\overrightarrow{L}} \right]\,x\,\overrightarrow{B}\,\,=\,\,0\left[ as\,\,\oint{d\,\overrightarrow{L}\,\,=\,\,0} \right]\)

        Further as area of coil,

                    \(\overrightarrow{S}\,\,=\,\,\left[ \frac{1}{3}\,\pi \,{{a}^{2}}\,\,-\,\,\frac{1}{2}\,.\,2a\,\sin {{60}^{0}}\,x\,a\,\cos {{60}^{0}} \right]\,\hat{k}\,\,=\,\,{{a}^{2}}\,\left[ \frac{\pi }{3}\,\,-\,\,\frac{\sqrt{3}}{4} \right]\hat{k}\)

        so        \(\overrightarrow{M}\) = \(I\,\overrightarrow{\,S\,}\,\,=\,\,I{{a}^{2}}\,\left[ \frac{\pi }{3}\,\,-\,\,\frac{\sqrt{3}}{4} \right]\,\hat{k}\)

and hence    \(\overrightarrow{\tau }\,\,=\,\,\overrightarrow{M}\,\,x\,\,\overrightarrow{B}\,\,=\,\,I\,{{a}^{2}}B\,\left[ \frac{\pi }{3}\,\,-\,\,\frac{\sqrt{3}}{4} \right]\,\left( \hat{k}\,x\hat{i} \right)\,\,\)

          i.e.          \(\overrightarrow{\tau }\,\,=\,\,I{{a}^{2}}\,B\,\,\left[ \frac{\pi }{3}\,\,-\,\,\frac{\sqrt{3}}{4} \right]\,\hat{j}\,N-mas\,\left( \hat{k}\,x\hat{i}\,\,=\,\hat{j} \right)\)

MAGNETISM

A solenoid is a long cylindrical helix, which is obtained by winding closely a large number of turns of insulated copper wire over a tube of cardboard or china clay. When electric current is passed through it, a magnetic field is produced around and within the solenoid.                                                                                                           

Consider a rectangular Amperean loop PQRS. Along PQ the field is zero. Along transverse sections PS and QR, the field component is zero. Let the magnetic field along SR be B.

                                ∮ Bdl cosθ = μ₀i

                                ∮ B→dl→ = ∮ Bdl cosθ

                                    = R∫S Bdl cos0 + Q∫R Bdl cos90^o + P∫Q Bdl cos180 + S∫P Bdl cos90^o

                                    = B R∫S dl + 0 + 0 + 0 = Bl

Current passing through loop – nli

Thus, the number of turns per unit length is ‘n’, then the total number of turns is ‘nh’. The enclosed current is Ie = I(nh) where I is the current in the solenoid.

From Ampere’s circuit law

                                                                      BL = μ₀I_e, Bh = μ₀I(nh)

                                                                      B = μ₀nI

The direction of the field is given by the right-hand rule. The solenoid is commonly used to obtain a uniform magnetic field.

1.         The magnetic field strength of a solenoid can be increased by inserting which material as the core?

Solution:

                   The Magnetic field of a solenoid increases when we insert a iron core because, iron is a ferromagnetic material and ferromagnetic                      materials help in increasing the magnetic property.

2.         What is the formula for magnetic field due to a solenoid?

Solution:

            The magnetic field due to a solenoid is:

          B= μnI, where μ is the permeability, n is the number of turns per unit length and I is the    current in the solenoid.

1.         A long solenoid has current I flowing through it, also denote N as the turns per unit length. Take its axis to be the z-axis, by                symmetry the only component of the magnetic field inside is Bz. Find the magnetic field at the center of the solenoid (on                      the axis).

Solution:

            Magnetic field at the center of a long solenoid is given by setting \({{\theta }_{1}}=\pi /2\And {{\theta }_{2}}=\pi \)

                                             \(B=\frac{{{\mu }_{0}}NI}{2}(\cos 0-\cos \pi )={{\mu }_{0}}NI\)

2.         A long solenoid has current I flowing through it, also denote N as the turns per unit length. Take its axis to be the z-axis, by                 symmetry the only component of the magnetic field inside is Bz. Find the magnetic field at the ends of the solenoid.

Solution:

            Magnetic field at the end of a long solenoid is given by setting  \({{\theta }_{1}}=\pi /2\And {{\theta }_{2}}=\pi \)

                                                      \(B=\frac{{{\mu }_{0}}NI}{2}(\cos \pi /2-\cos \pi )=\frac{{{\mu }_{0}}NI}{2}\)

1.            Find the magnetic field produced by the solenoid if the number of loop is 400 and current passing through on it is 5 A.                          (Length of the solenoid is 40cm and k= \({{10}^{-7}}\) \(N/Amp{{s}^{-2}}\))

           N=400, i=5A, l=40cm=0.4m, k= \({{10}^{-7}}\) \(N/Amp{{s}^{-2}}\))

Solution:

                      \(\begin{align} & B=4\pi k\frac{i.N}{l}=4.3\frac{{{10}^{-7}}.5.400}{0.4} \\ & B={{6.10}^{-3}}\frac{N}{Amps.m} \\ \end{align}\)

2.         A solenoid has 80 cm diameter, number of loop is 4 and magnetic field inside it is 1,2.10^-5N/Amp.m. Find the current passing               through the each loop of wire. Since the questions ask current on each loop, we assume each loop as circle thus we find the               magnetic field

Solution:

                     \(\begin{align} & {{i}_{loop}}=n.i \\ & r=\frac{80}{2}=40cm=0.4m \\ & B=1,{{2.10}^{-4}}\frac{N}{Amps.m} \\ & n=4 \\ & B={{2.3.10}^{-7}}.\frac{4.i}{0.4} \\ & i=2Amps \\ \end{align}\)

1.         In the figure shown a coil of single turn is wound on a sphere of radius r and mass m. The plane of the coil is parallel to the                inclined plane and lies in the equatorial plane of the sphere. If sphere is in rotational equilibrium the value of B is (current in                the coil is i)

Solution:      

2.         A charged particle P leaves the origin with speed v = v0, at some inclination with the x-axis. There is a uniform magnetic field              B along the x-axis. P strikes a fixed target T on the x-axis for a minimum value of B = B0. P will also strike T if

Solution:

               Let d = distance of the target T from the point of projection. P will strike T if d an integral

               multiple of the pitch.

                                                 Pitch = \(\left( 2\pi \frac{m}{QB} \right)\ v\ \cos \theta \)

                Here,  m, Q and q are constant

                                     \      pitch = \(k\left( \frac{v}{B} \right)\) where k = constant

                                                Initially, d = k \(\left( \frac{{{v}_{0}}}{{{B}_{0}}} \right)\)

MAGNETISM

A solenoid is a long cylindrical helix, which is obtained by winding closely a large number of turns of insulated copper wire over a tube of cardboard or china clay. When electric current is passed through it, a magnetic field is produced around and within the solenoid.                                                                                                           

Consider a rectangular Amperean loop PQRS. Along PQ the field is zero. Along transverse sections PS and QR, the field component is zero. Let the magnetic field along SR be B.

                                ∮ Bdl cosθ = μ₀i

                                ∮ B→dl→ = ∮ Bdl cosθ

                                    = R∫S Bdl cos0 + Q∫R Bdl cos90^o + P∫Q Bdl cos180 + S∫P Bdl cos90^o

                                    = B R∫S dl + 0 + 0 + 0 = Bl

Current passing through loop – nli

Thus, the number of turns per unit length is ‘n’, then the total number of turns is ‘nh’. The enclosed current is Ie = I(nh) where I is the current in the solenoid.

From Ampere’s circuit law

                                                                      BL = μ₀I_e, Bh = μ₀I(nh)

                                                                      B = μ₀nI

The direction of the field is given by the right-hand rule. The solenoid is commonly used to obtain a uniform magnetic field.

1.         The magnetic field strength of a solenoid can be increased by inserting which material as the core?

Solution:

                   The Magnetic field of a solenoid increases when we insert a iron core because, iron is a ferromagnetic material and ferromagnetic                      materials help in increasing the magnetic property.

2.         What is the formula for magnetic field due to a solenoid?

Solution:

            The magnetic field due to a solenoid is:

          B= μnI, where μ is the permeability, n is the number of turns per unit length and I is the    current in the solenoid.

1.         A long solenoid has current I flowing through it, also denote N as the turns per unit length. Take its axis to be the z-axis, by                symmetry the only component of the magnetic field inside is Bz. Find the magnetic field at the center of the solenoid (on                      the axis).

Solution:

            Magnetic field at the center of a long solenoid is given by setting \({{\theta }_{1}}=\pi /2\And {{\theta }_{2}}=\pi \)

                                             \(B=\frac{{{\mu }_{0}}NI}{2}(\cos 0-\cos \pi )={{\mu }_{0}}NI\)

2.         A long solenoid has current I flowing through it, also denote N as the turns per unit length. Take its axis to be the z-axis, by                 symmetry the only component of the magnetic field inside is Bz. Find the magnetic field at the ends of the solenoid.

Solution:

            Magnetic field at the end of a long solenoid is given by setting  \({{\theta }_{1}}=\pi /2\And {{\theta }_{2}}=\pi \)

                                                      \(B=\frac{{{\mu }_{0}}NI}{2}(\cos \pi /2-\cos \pi )=\frac{{{\mu }_{0}}NI}{2}\)

1.            Find the magnetic field produced by the solenoid if the number of loop is 400 and current passing through on it is 5 A.                          (Length of the solenoid is 40cm and k= \({{10}^{-7}}\) \(N/Amp{{s}^{-2}}\))

           N=400, i=5A, l=40cm=0.4m, k= \({{10}^{-7}}\) \(N/Amp{{s}^{-2}}\))

Solution:

                      \(\begin{align} & B=4\pi k\frac{i.N}{l}=4.3\frac{{{10}^{-7}}.5.400}{0.4} \\ & B={{6.10}^{-3}}\frac{N}{Amps.m} \\ \end{align}\)

2.         A solenoid has 80 cm diameter, number of loop is 4 and magnetic field inside it is 1,2.10^-5N/Amp.m. Find the current passing               through the each loop of wire. Since the questions ask current on each loop, we assume each loop as circle thus we find the               magnetic field

Solution:

                     \(\begin{align} & {{i}_{loop}}=n.i \\ & r=\frac{80}{2}=40cm=0.4m \\ & B=1,{{2.10}^{-4}}\frac{N}{Amps.m} \\ & n=4 \\ & B={{2.3.10}^{-7}}.\frac{4.i}{0.4} \\ & i=2Amps \\ \end{align}\)

1.         In the figure shown a coil of single turn is wound on a sphere of radius r and mass m. The plane of the coil is parallel to the                inclined plane and lies in the equatorial plane of the sphere. If sphere is in rotational equilibrium the value of B is (current in                the coil is i)

Solution:      

2.         A charged particle P leaves the origin with speed v = v0, at some inclination with the x-axis. There is a uniform magnetic field              B along the x-axis. P strikes a fixed target T on the x-axis for a minimum value of B = B0. P will also strike T if

Solution:

               Let d = distance of the target T from the point of projection. P will strike T if d an integral

               multiple of the pitch.

                                                 Pitch = \(\left( 2\pi \frac{m}{QB} \right)\ v\ \cos \theta \)

                Here,  m, Q and q are constant

                                     \      pitch = \(k\left( \frac{v}{B} \right)\) where k = constant

                                                Initially, d = k \(\left( \frac{{{v}_{0}}}{{{B}_{0}}} \right)\)