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1.	The wires which connect the battery of an automobile to its starting motor carry a current of `300A` (for a short time). What is the force per unit length between the wires if they are `70cm` long and `1*5cm` apart? Is the force attractive or repulsive?
Answer» Current in both wires, `I = 300A` Distance between the wires, `r = 1.5 cm = 0.015m` Length of the wires, `l = 70 cm = 0.7m` Force between the two wires is given by the relation, `F = (mu_(0)I^(2))/(2pir)` Where, `mu_(0)=` Permeability of free space `= 4pi xx 10^(-7)TmA^(-1)` `:. F = (4pi xx 10^(-7) xx (300)^(2))/(2pi xx 0.015)` `=1.2 N//m` Since the direction of the circuit in the wires is opposite, a repulsive force exists between them.

2.	Two moving coil meters `M_1` and `M_2` have the following particulars: `R_1=10Omega`, `N_1=30`, `A_1=36xx10^-3m^2`, `B_1=025T`, `R_2=14Omega`, `N_2=42`, `A_2=18xx10^-3m^2`, `B_2=050T` (The spring constants are identical for the two metres). Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of `M_2` and `M_1`.
Answer» For moving coil meter `M_(1)`: Resistance, `R_(1) = 10 Omega` Number of turns, `N_(1) = 30` Area of cross-section, `A_(1) = 3.6 xx 10^(-3)m^(2)` Magnetic field strength, `B_(1) = 0.25T` Spring constant `K_(1) = K` For moving coil meter `M_(2)`: Resistance, `R_(2) = 14 Omega` Number of turns, `N_(2) = 42` Area of cross-section, `A_(2) = 1.8 xx 10^(-3)m^(2)` Magnetic field strength, `B_(2) = 0.50T` Spring constant, `K_(2) = K` (a) Current sensitivity of `M_(1)` is given as: `I_(s1) = (N_(1)B_(1)A_(1))/(K_(1))` And, current sensitivity of `M_(2)` is gives as: `I_(s2) = (N_(2)B_(2)A_(2))/(K_(2))` `(I_(s2))/(I_(s1)) = (N_(2)B_(2)A_(2)K_(1))/(K_(2)N_(1)B_(1)A_(1))` `:.` Ratio `=(42 xx 0.5 xx 1.8 xx 10^(-3)xxK)/(K xx 30 xx 0.25 xx 3.6 xx 10^(-3)) = 1.4` Hence, the ratio of current sensitivity of `M_(2)` to `M_(1)` is 1.4. (b) Voltage senstivity for `M_(2)` is given as: `V_(s2) = (N_(2)B_(2)A_(2))/(K_(2)R_(2))` And, voltage sensitivity for `M_(1)` is given as: `V_(s1) = (N_(1)B_(1)A_(1))/(K_(1))` `(V_(s2))/(V_(s1)) = (N_(2)B_(2)A_(2)K_(1)R_(1))/(K_(2)R_(2)N_(1)B_(1)A_(1))` `:.` Ratio `= (42 xx 0.5 xx 1.8 xx 10^(-3)xx 10 xx K)/(Kxx 14 xx 30 xx 0.25 xx 3.6 xx 10^(-3)) =1` Hence, the ratio of voltage sensitivity of `M_(2)` to `M_(2)` to 1.

3.	A uniform magnetic field of `3000G` is established along the positive z-direction. A rectangular loop of sides `10cm` and `5cm` carries a current `12A`. What is the torque on the loop in the different cases shown in the figure. What is the force on each case? Which case corresponds to stable equilibrium?
Answer» Magnetic field strength, `B = 3000G = 3000 xx 10^(-4)T = 0.3T` Length of the rectangular loop, `l = 10cm` Width of the rectangular loop, `b = 5cm` Area of the loop, ` A= l xx b = 10 xx 5 = 50 cm^(2) = 50 xx 10^(-4)m^(2)` Current in the loop, `I = 12A` Now, taking the anti-clockwise direction of the current as positive and vise-versa: (a) Torque, `vec(tau) = I vec(A) xx vec(B)` From the given figure, it can be observed that A is normal to the y-z plane and B is directed along the z-axis. `:. r = 12 xx (50 xx 10^(-4)) hati xx 0.3 hatk` `=-0 1.8 xx 10^(-2)hatj Nm` The torque is `1.8 xx 10^(-2)Nm` along the negative y-direction. The force on the loop is zero because the angle between A and B is zero. (b) This case is similar to case (a). Hence, the answer is the same as(a). (c) Torque `tau = IA xx B` From the given figure, it can be observed that A is normal to the x-z plane and B is directed along the z-axis. `:. tau = - 12 xx (50 xx 10^(-4))hatj xx 0.3 hatk` `=- 1.8 xx 10^(-2)hatj Nm` The torque is `1.8 xx 10^(-2)Nm` along the negative x direction and the force is zero. (d) Magnitude of torque is given as: `\|tau\| = IAB` `= 12 xx 50 xx 10^(-4) xx 0.3` `= 1.8 xx 10^(-2)Nm` ltbr. Torque is `1.8 xx 10^(-2)Nm` at an angle of `240^(@)` with positive x direction. The force is zero. (e) Torque `tau = l vec(A) xx vec(B)` `= (50 xx 10^(-4) xx 12) hatk xx 0.3 hatk = 0` Hence, the torque is zero. The force is also zero. (f) torque `tau = l vec(a) xx vec(B)` `=(50 xx 10^(-4) xx 12) hatk xx 0.3 hatk =0` Hence, the torque is zero. The force is also zero. In case (e), The direction of `I vec(A)` and `vec(B)` is the same and the angle between them is zero. If displaced, they come back to an equilibrium. hence, its equilibrium is stable. Whereas, in case (f), the direction of `I vec(A)` and `vec(B)` is opposite. The angle between them is `180^(@)`. If disturbed, it does not come back to its original position. Hence, its equilibrium is unstable.

4.	A square coil of side `10cm` consists of 20 turns and carries a current of 12A. The coil is suspended vertically and normal to the plane of the coil makes an angle of `30^@` with the direction of a uniform horizontal magnetic field of magnitude `0*80T`. What is the magnitude of torque experienced by the coil?
Answer» Length of a side of the square coil, `l = 10 cm = 0.1m` Current flowing in the coil, `I = 12A` Number of turns on the coil, `n = 20` Angle made by the plane of the coil with magnetic field, `theta = 30^(@)` Strength of magnetic field, `B = 0.80T` Magnitude of the magnetic torque experienced by the coil in the magnetic field is given by the relation, `T = n BIA sin theta` Where, `A =` Area of the square coil `rArr l xxl = 0.1 xx 0.1 = 0.01 m^(2)` `:. T = 20 xx 0.8 xx 12 xx 0.01 xx sin 30^(@)` `= 0.96N` Hence, the magnitude of the torque experienced by the coil is `0.96Nm`.

5.	A straight wire carrying a current of `12A` is bent into a semicircular arc of radius `2*0cm` as shown in figure. Consider the magnetic field `vecB` at the centre of arc. (a) What is the magnetic field due to the staight segments? (b) In what way the contribution to `vecB` from the semicircle differs from that of a circular loop and in what way does it resemble? (c) Would your answer be different if the wire were bent into a semicircle arc of the same radius but in the opposite way as shown in figure
Answer» (a) `dl` and r for each element of the straight segments are parallel. Therefore, `dl xx r = 0`. Straight segments do not contribute to \|B\|. (b) For all segments of the semicircular arc, `dl x× r` are all parallel to each other (into the plane of the paper). All such contributions add up in magnitude. Hence direction of B for a semicircular arc is given by the right-hand rule and magnitude is half that of a circular loop. Thus B is `1.9 xx 10^(–4) T` normal to the plane of the paper going into it. (c) Same magnitude of B but opposite in direction to that in (b).

6.	A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is suspended in mid-air by a uniform horizontal magnetic field B. What is the magnitude of the magnetic field?
Answer» From Eq. (4.4), we find that there is an upward force F, of magnitude IlB,. For mid-air suspension, this must be balanced by the force due to gravity: `m g = I lB` `B = (mg)/(Il)` `=(0.2 xx 9.8)/(2 xx 1.5) = 0.65 T` Note that it would have been sufficient to specify `m//l`, the mass per unit length of the wire. The earth’s magnetic field is approximately `4 ×x 10^(–5) T` and we have ignored it.

7.	What is the radius of the path of an electron (mass `9 xx 10^(-31)` kg and charge `1.6 xx 10^(-19)C)` moving at a speed of `3 xx 10^(7)m//s` in a magnetic field of `6 xx 10^(-4)T` perpendicular to it? What is its frequency? Calculate its energy in `keV.(1eV = 1.6 xx 10^(-19)J)`.
Answer» Using Eq. (4.5) we find `r = mv//(qB) = 9 xx 10^(-31) kg xx 3 xx 10^(7)ms^(-1)//(1.6 xx 10^(-19)C xx 6 xx 10^(-4)T)` `= 26 xx 10^(-2)m = 26 cm` `v = v//(2pir) = 2 xx 10^(6)s^(-1) = 2xx 10^(6)Hz = 2 MHz`. `E = (½) mv^(2) = (½) 9 xx 10^(-31)kg xx 9 xx 10^(14)m^(2)//s^(2) = 40.5 xx 10^(-17)J` `~~4 xx 10^(-16)J = 2.5 keV`.

8.	If the magnetic field is parallel to the positive y-axis and the charged particle is moving along the positive x-axis (Fig.), which way would the Lorentz force be for (a) an electron (negative charge), (b) a proton (positive charge).
Answer» The velocity v of particle is along the x-axis, while B, the magnetic field is along the y-axis, so `v x× B` is along the z-axis (screw rule or right-hand thumb rule). So, (a) for electron it will be along `–z` axis. (b) for a positive charge (proton) the force is along `+z` axis.

9.	A current carrying circular loop is located in a uniform external magnetic field. If the loop if free to turn, what is its orientation of stable equilibrium? Show that in this orientation, the flux of the total field (external field + field produced by the loop) is maximum.
Answer» (a) No, because that would require t to be in the vertical direction. But `tau = I A x× B`, and since A of the horizontal loop is in the vertical direction, t would be in the plane of the loop for any B. (b) Orientation of stable equilibrium is one where the area vector A of the loop is in the direction of external magnetic field. In thisorientation, the magnetic field produced by the loop is in the same direction as external field, both normal to the plane of the loop, thus giving rise to maximum flux of the total field. (c) It assumes circular shape with its plane normal to the field to maximise flux, since for a given perimeter, a circle encloses greater area than any other shape.

10.	In the circuit shown in figure, the current is to measured. What is the value of the current if the ammeter shown (i) is a galvanometer with a resistance `R_G=6000Omega`. (ii) is a galvanometer described in (i) but converted to an ammeter by a shunt resistance `r_s=002Omega`, (iii) is an ideal ammeter with zero resistance?
Answer» (a) Total resistance in the circuit is, `R_(G) +3 = 63 Omega`. Hence, `I = 3//63 = 0.048A`. (b) Resistance of the galvanometer converted to an ammeter is, `(R_(G)r_(s))/(R_(G)+r_(s)) = (60 Omega xx 0.02 Omega)/((60+0.02)Omega) ~~ 0.02Omega` Total resistance in the circuit is, `0.02 Omega + 3 Omega = 3.02 Omeg`. Hence, `I = 3//3.02 = 0.99A`. (c) For the ideal ammeter with zero resistance, `I = 3//3 = 1.00A`