A multirange voltmeter can be constructed by using a galvanometer circuit as shown in figure. We want to construct a voltmeter that can measure $2 \mathrm{~V}, 20 \mathrm{~V}$ and 200 V using a galvanometer of resistance $10 \Omega$ and that produces maximum deflection for current of 1 mA . Find $R_1, R_2$ and $R_3$ that have to be used.
Applying expression in different situations
For $i_G\left(G+R_1\right)=2$ for 2 V range
For $i_G\left(G+R_1+R_2\right)=20$ for 20 V range
and For $i_G\left(G+R_1+R_2+R_3\right)=200$ for 200 V range
On solving, we get $R_1=1990 \Omega, R_2=18 \mathrm{k} \Omega$ and $R_3=180 \mathrm{k} \Omega$.
A long straight wire carrying current of 25 A rests on a table as shown in figure. Another wire $P Q$ of length 1 m , mass 2.5 g carries the same current but in the opposite direction. The wire $P Q$ is free to slide up and down. To what height will $P Q$ rise?
The magnetic field produced by long straight wire carrying current of 25 A rests on a table on small wire
$$B=\frac{\propto_0 I}{2 \pi h}$$
The magnetic force on small conductor is
$$F=B I l \sin \theta=B I l$$
Force applied on $P Q$ balance the weight of small current carrying wire.
$$\begin{aligned} & F=m g=\frac{\propto_0 I^2 l}{2 \pi h} \\ & h=\frac{\propto_0 I^2 l}{2 \pi m g}=\frac{4 \pi \times 10^{-7} \times 25 \times 25 \times 1}{2 \pi \times 2.5 \times 10^{-3} \times 9.8}=51 \times 10^{-4} \\ & h=0.51 \mathrm{~cm} \end{aligned}$$
A 100 turn rectangular coil $A B C D$ (in $X-Y$ plane) is hung from one arm of a balance figure. A mass 500 g is added to the other arm to balance the weight of the coil. A current 4.9 A passes through the coil and a constant magnetic field of 0.2 T acting inward (in $x-z$ plane) is switched on such that only arm CD of length 1 cm lies in the field. How much additional mass $m$ must be added to regain the balance?
For equilibrium/ balance, net torque should also be equal to zero.
When the field is off $\sum t=0$ considering the separation of each hung from mid-point be $I$.
$$\begin{aligned} M g l & =W_{\text {coil }} l \\ 500 \mathrm{gl} & =W_{\text {coil }} l \\ W_{\text {coil }} & =500 \times 9.8 \mathrm{~N} \end{aligned}$$
Taking moment of force about mid-point, we have the weight of coil
When the magnetic field is switched on
$$\begin{aligned} M g l+m g l & =W_{\text {coil }} l+I B L \sin 90 Y I \\ m g l & =\text { BIL } l \\ m & =\frac{B I L}{g}=\frac{0.2 \times 4.9 \times 1 \times 10^{-2}}{9.8}=10^{-3} \mathrm{~kg}=1 \mathrm{~g} \end{aligned}$$
Thus, 1 g of additional mass must be added to regain the balance.
A rectangular conducting loop consists of two wires on two opposite sides of length $l$ joined together by rods of length $d$. The wires are eachof the same material but with cross-sections differing by a factor of 2 . The thicker wire has a resistance $R$ and the rods are of low resistance, which in turn are connected to a constant voltage source $V_0$. The loop is placed in uniform a magnetic field $\mathbf{B}$ at $45^{\circ}$ to its plane. Find $\tau$, the torque exerted by the magnetic field on the loop about an axis through the centres of rods.
The thicker wire has a resistance $R$, then the other wire has a resistance $2 R$ as the wires are of the same material but with cross-sections differing by a factor 2.
Now, the force and hence, torque on first wire is given by
$$F_1=i_1 l B=\frac{V_0}{2 R} l B, \tau_1=\frac{d}{2 \sqrt{2}} F_1=\frac{V_0 l d B}{2 \sqrt{2} R}$$
Similarly, the force hence torque on other wire is given by
$$F_2=i_2 l B=\frac{V_0}{2 R} l B, \tau_2=\frac{d}{2 \sqrt{2}} F_2=\frac{V_0 l d B}{4 \sqrt{2} R}$$
$$\begin{aligned} &\text { So, net torque, }\\ &\begin{aligned} \tau & =\tau_1-\tau_2 \\ \tau & =\frac{1}{4 \sqrt{2}} \frac{V_0 l d B}{R} \end{aligned} \end{aligned}$$
An electron and a positron are released from $(0,0,0)$ and $(0,0,1.5 R)$ respectively, in a uniform magnetic field $\mathbf{B}=B_0 \hat{\mathbf{i}}$, each with an equal momentum of magnitude $p=e B R$. Under what conditions on the direction of momentum will the orbits be non-intersecting circles?
Since, $B$ is along the $x$-axis, for a circular orbit the momenta of the two particles are in the $y$-z plane. Let $p_1$ and $p_2$ be the momentum of the electron and positron, respectively. Both traverse a circle of radius $R$ of opposite sense. Let $p_1$ make an angle $\theta$ with the $y$-axis $p_2$ must make the same angle.
The centres of the respective circles must be perpendicular to the momenta and at a distance $R$. Let the centre of the electron be at $C_e$ and of the positron at $C_p$. The coordinates of $C_e$ is
$$C_e \equiv(0,-R \sin \theta, R \cos \theta)$$
The coordinates of $C_p$ is
$$C_p \equiv\left(0,-R \sin \theta, \frac{3}{2} R-R \cos \theta\right)$$
The circles of the two shall not overlap if the distance between the two centers are greater than $2 R$.
Let $d$ be the distance between $C_p$ and $C_e$.
Let $d$ be the distance between $C_p$ and $C_e$.
Then,
$$\begin{aligned} d^2 & =(2 R \sin \theta)^2+\left(\frac{3}{2} R-2 R \cos \theta\right)^2 \\ & =4 R^2 \sin ^2 \theta+\frac{9^2}{4} R-6 R^2 \cos \theta+4 R^2 \cos ^2 \theta \\ & =4 R^2+\frac{9}{4} R^2-6 R^2 \cos \theta \end{aligned}$$
Since, $d$ has to be greater than $2 R$
$$\begin{array}{rlrl} & d^2 >4 R^2 \\ \Rightarrow & 4 R^2+\frac{9}{4} R^2-6 R^2 \cos \theta >4 R^2 \\ \Rightarrow & \frac{9}{4} >6 \cos \theta \\ \text { or, } & \cos \theta <\frac{3}{8} \end{array}$$