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In a moving coil galvanometer the coil is suspended between the pole pieces of a strong horse-shoe magnet. The pole pieces are made cylindrical and a soft iron cylindrical core is placed within the coil without touching it. This makes the field radial. In such a field the plane of the coil always remains parallel to the field. Therefore \[\theta ={{90}^{o}}\] and the deflecting torque always has the maximum value. \[{{\tau }_{\text{def}}}=NBiA\]                 ......(i) Coil deflects, a restoring torque is set up in the suspension fibre. If \[\alpha \] is the angle of twist, the restoring torque is \[{{\tau }_{\text{rest}}}=C\alpha \]                                         .....(ii) where C is the torsional constant of the fibre. When the coil is in equilibrium \[NBiA=C\alpha \Rightarrow i=Ka,\] where \[K=\frac{C}{NBA}\] is the galvanometer constant. This linear relationship between i and a makes the moving coil galvanometer useful for current measurement and detection. Current sensitivity \[\mathbf{(}{{\mathbf{S}}_{\mathbf{i}}}\mathbf{)}\] : The current sensitivity of a galvanometer is defined as the deflection produced in the galvanometer per unit current flowing through it. \[{{S}_{i}}=\frac{\alpha }{i}=\frac{NBA}{C}\] Voltage sensitivity \[\mathbf{(}{{\mathbf{S}}_{\mathbf{V}}}\mathbf{)}\] : Voltage sensitivity of a galvanometer is defined as the deflection produced in the galvanometer per unit voltage applied to it. \[{{S}_{V}}=\frac{\alpha }{V}=\frac{\alpha }{iR}=\frac{{{S}_{i}}}{R}=\frac{NBA}{RC}\]  

(1) Torque : Consider a rectangular current carrying coil PQRS having N turns and area A, placed in a uniform field \[\overrightarrow{B},\] in such a way that the normal \[(\hat{n})\] to the coil makes an angle \[\theta \] with the direction of \[\overrightarrow{B}\]. The coil experiences a torque given by \[\tau =NBiA\sin \theta \]. Vectorially \[\overrightarrow{\tau }=\overrightarrow{M}\times \overrightarrow{B}\] (i) \[\tau \] is zero when \[\theta =0,\] i.e., when the plane of the coil is perpendicular to the field. (ii) \[\tau \] is maximum when \[\theta ={{90}^{o}}\], i.e., the plane of the coil is parallel to the field  \[{{\tau }_{\max }}=NBiA\] (2) Workdone : If coil is rotated through an angle \[\theta \] from it's equilibrium position then required work. \[W=MB(1-\cos \theta ).\] It is maximum when \[\theta ={{180}^{o}}\Rightarrow {{W}_{\max }}=2MB\] (3) Potential energy :  \[U=-MB\cos \theta \Rightarrow U=-\overrightarrow{M}.\overrightarrow{B}\]

A current carrying circular coil behaves as a bar magnet whose magnetic moment is M = NiA; Where N = Number of turns in the coil, i = Current through the coil  and A = Area of the coil Magnetic moment of a current carrying coil is a vector and it's direction is given by right hand thumb rule (1) For a given perimeter circular shape have maximum area. Hence maximum magnetic moment. (2) For a any loop or coil \[\overrightarrow{B}\] at centre due to current in loop, and \[\overrightarrow{M}\] are always parallel.

Case 1 : When an arbitrary current carrying loop placed in a magnetic field (\[\bot \]to the plane of loop), each element of loop experiences a magnetic force due to which loop stretches and open into circular loop and tension developed in it's each part. Case 2 : Equilibrium of a current carrying conductor : When a finite length current carrying wire is kept parallel to another infinite length current carrying wire, it can suspend freely in air as shown below In both the situations for equilibrium of XY it's downward weight = upward magnetic force i.e. \[\mathbf{mg=}\frac{{{\mathbf{\mu }}_{\mathbf{0}}}}{\mathbf{4\pi }}\mathbf{.}\frac{\mathbf{2}{{\mathbf{i}}_{\mathbf{1}}}{{\mathbf{i}}_{\mathbf{2}}}}{\mathbf{h}}\mathbf{.l}\] Case 3 : Current carrying spring : If current is passed through a spring, then it will contract because current will flow through all the turns in the same direction. Case 4 : Tension less strings : In the following figure the value and direction of current through the conductor XY so that strings becomes tensionless? Strings becomes tensionless if weight of conductor XY balanced by magnetic force \[({{F}_{m}})\]. Hence direction of current is from \[X\to Y\] and in balanced condition \[{{F}_{m}}=mg\]  \[\Rightarrow \] \[B\,i\,l=mg\] \[\Rightarrow \] \[i=\frac{mg}{B\,l}\] Case 5 : Sliding of conducting rod on inclined rails : When a conducting rod slides on conducting rails. In the following situation conducting rod (X, Y) slides at constant velocity if \[F\cos \theta =mg\sin \theta \]\[\Rightarrow \]\[B\,i\,l\cos \theta =mg\sin \theta \]\[\Rightarrow \]\[B=\frac{mg}{i\,l}\tan \theta \]

If two charges \[{{q}_{1}}\] and \[{{q}_{2}}\] are moving with velocities v1 and v2 respectively and at any instant the distance between them is r, then . Magnetic force between them is \[{{F}_{m}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{{{q}_{1}}{{q}_{2}}\,{{v}_{1}}{{v}_{2}}}{{{r}^{2}}}\] .... (i) and Electric force between them is \[{{F}_{e}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\]  .... (ii) From equation (i) and (ii) \[\frac{{{F}_{m}}}{{{F}_{e}}}={{\mu }_{0}}{{\varepsilon }_{0}}{{v}^{2}}\] but \[{{\mu }_{0}}{{\varepsilon }_{0}}=\frac{1}{{{c}^{2}}}\];  where c is the velocity of light in vacuum. So \[\frac{{{F}_{m}}}{{{F}_{e}}}={{\left( \frac{v}{c} \right)}^{2}}\] As \[v<c\] so \[{{F}_{m}}<{{F}_{e}}\]

The force on a length l of each of two long, straight, parallel wires carrying currents \[{{i}_{1}}\] and \[{{i}_{2}}\] and separated by a distance a is \[F=\frac{{{\mu }_{0}}}{4\pi }\cdot \frac{2{{i}_{1}}{{i}_{2}}}{a}\times l\] Hence force per unit length \[\frac{F}{l}=\frac{{{\mu }_{0}}}{4\pi }\cdot \frac{2{{i}_{1}}{{i}_{2}}}{a}\]\[\left( \frac{N}{m} \right)\]or\[\frac{F}{l}=\frac{2{{i}_{1}}{{i}_{2}}}{a}\left( \frac{dyne}{cm} \right)\] Direction of force : If conductors carries current in same direction, then force between them will be attractive. If conductor carries current in opposite direction, then force between them will be repulsive.  

If a particle carrying a positive charge q and moving with velocity v enters a magnetic field B then it experiences a force F which is given by the expression \[\overrightarrow{F}=q(\overrightarrow{v\,}\times \overrightarrow{B})\] \[\Rightarrow \] \[F=qvB\sin \theta \] where \[\overrightarrow{v\,}\,=\] velocity of the particle, \[\overrightarrow{B}=\] magnetic field (1) Zero force : Force on charged particle will be zero (i.e. \[F=0\]) if (i) No field i.e. \[B=0\Rightarrow F=0\] (ii) Neutral particle i.e. \[q=0\,\Rightarrow F=\text{ }0\] (iii) Rest charge i.e. \[v=0\,\Rightarrow F=0\] (iv) Moving charge i.e. \[\theta ={{0}^{o}}\]or \[\theta ={{180}^{o}}\Rightarrow F=0\] (2) Direction of force : The force \[\overrightarrow{F}\] is always perpendicular to both the velocity \[\overrightarrow{v}\] and the field \[\overrightarrow{B}\] in accordance with Right Hand Screw Rule, though \[\overrightarrow{v}\] and \[\overrightarrow{B}\] themselves may or may not be perpendicular to each other. Direction of force on charged particle in magnetic field can also be find by Fleming's Left Hand Rule (FLHR). Here, First finger (indicates) \[\to \]Direction of magnetic field Middle finger \[\to \] Direction of motion of positive charge or direction, Opposite to the motion of negative charge. Thumb \[\to \] Direction of force

A toroid can be considered as a ring shaped closed solenoid. Hence it is like an endless cylindrical solenoid. Consider a toroid having n turns per unit length. Magnetic field at a point P in the figure is given as  \[B=\frac{{{\mu }_{0}}Ni}{2\pi r}={{\mu }_{o}}ni\] where \[n=\frac{N}{2\pi r}\]  

A cylinderical coil of many tightly wound turns of insulated wire with generally diameter of the coil smaller than its length is called a solenoid. A magnetic field is produced around and within the solenoid. The magnetic field within the solenoid is uniform and parallel to the axis of solenoid. (1) Finite length solenoid : If \[N=\] total number of turns, \[l=\] length of the solenoid, \[n=\] number of turns per unit length \[=\frac{N}{l}\] (i) Magnetic field inside the solenoid at point P is given by  \[B=\frac{{{\mu }_{0}}}{4\pi }(2\pi \,ni)[\sin \alpha +\sin \beta ]\] (ii) Infinite length solenoid : If the solenoid is of infinite length and the point is well inside the solenoid i.e. \[\alpha =\beta =(\pi /2)\]. So           \[{{B}_{in}}={{\mu }_{0}}ni\] (iii) If the solenoid is of infinite length and the point is near one end i.e. \[\alpha =0\] and \[\beta =(\pi /2)\] so \[{{B}_{end}}=\frac{1}{2}\mathbf{(}{{\mu }_{\mathbf{0}}}ni\mathbf{)}\]       (\[{{B}_{end}}=\frac{1}{2}{{B}_{in}}\])  

The figure shows an infinite sheet of current with linear current density j (A/m). Due to symmetry the field line pattern above and below the sheet is uniform. Consider a square loop of side l as shown in the figure. \[\int_{a}^{b}{\overrightarrow{B}.\overrightarrow{dl}}+\int_{b}^{c}{\overrightarrow{B}.\overrightarrow{dl}}+\int_{c}^{d}{\overrightarrow{B}.\overrightarrow{dl}}+\int_{d}^{a}{\overrightarrow{B}.\overrightarrow{dl}}={{\mu }_{0}}i\]   (By Ampere's law) Since \[B\bot dl\] along the path \[b\to c\] and \[d\to a,\] therefore, \[\int_{b}^{c}{\overrightarrow{B}.\overrightarrow{dl}}=0\]; \[\int_{d}^{a}{\overrightarrow{B}.\overrightarrow{dl}}=0\] Also, \[B\,\,|\,\,|\,\,dl\] along the path \[a\to b\] and \[c\to d,\] thus \[\int_{a}^{b}{\overrightarrow{B}.\overrightarrow{dl}}+\int_{d}^{a}{\overrightarrow{B}.\overrightarrow{dl}}=2Bl\] The current enclosed by the loop is \[i=jl\]. Therefore, according to Ampere?s law \[2Bl={{\mu }_{0}}(jl)\] or \[B=\frac{{{\mu }_{0}}j}{2}\]