Question 1

A current  \mathrm{I}  flows in a infinitely long wire with cross section in the form of a semicircular ring of radius R. The magnitude of the magnetic induction at its axis is

Accepted Answer

In the figure we have shown the cross section (of the given wire) lying in the XY plane. The length of the wire is along the Z-axis and the current in the wire is supposed to flow along the negative Z-direction. The broad infinitely long wire can be imagined to be made of a large number of infinitely long straight wire strips, each of small width  dl .With reference to the figure, we have  dl = Rd \, 	heta .The magnetic flux density due to the above strip is shown as  dB_1  in the figure. It has an X-component  dB_1 \, sin \, 	heta  and Y-component  dB_1\, cos \, 	heta . When we consider a similar strip of the same with  dl  located symmetrically with respect to the Y-axis, we obtain a contribution  dB_2  to the flux density. The flux density  dB_2  has the same magnitude as  dB_1 . It has X-component  dB_2 \, sin \, 	heta  and Y-component  dB_2 \,cos \, 	heta . The X-components of  dB_1  and  dB_1  are in the same magnitude and direction and they add up. But the Y-components of  dB_1  and  dB_1  are in opposite directions and have the same magnitude. Therefore they get canceled. The entire conductor therefore produces a resultant magnetic field along the negative X-direction.The wire strip of width  dl  can be imagined to be an ordinary thin straight infinitely long wire carrying current  \frac{Idl }{ \pi R}  since the total current I flows through the semicircular cross section of perimeter  \pi R . Putting  dB_1 = dB_2 = dB  we have \displaystyle dB = \mu\frac{\frac{Idl}{\pi R}}{2\pi R} = \mu_0\frac{ \frac{IRd	heta}{\pi R}}{2\pi R} = \frac{\mu_0Id	heta}{2\pi^2R} The X-component of the above field is  \displaystyle \frac{\mu_0I}{2\pi^2R}sin \, 	heta \, d\, 	heta The field due to the entire conductor is  B = \int _0 ^{\pi} [\frac{\mu_0I}{2\pi^2R}sin \, 	heta]d\, 	heta or,  B = \frac{\mu_0I}{\pi^2R}  since  \int _0 ^{\pi} sin \, 	heta d 	heta = 2

Question 2

A long straight wire of radius  a  carries a steady current i. The current is uniformly distributed across its cross section. The ratio of the magnetic field at  a/2  and  2a  is:

Accepted Answer

Here, current is uniformly distributed across the cross-section of the wire, therefore, current enclosed in the amperean path formed at a distance  r_1=\displaystyle\left(\frac{a}{2}ight) I_{in}=\displaystyle\left(\frac{\pi r^2_1}{\pi a^2}ight)	imes I , where  I  is total current. 	herefore  Magnetic field at  (B_1)=\displaystyle\frac{\mu_0	imes current \ inclosed}{path}             =\displaystyle\frac{\displaystyle\mu_0	imes \left(\frac{\pi r^2_1}{\pi a^2}ight)	imes I}{2\pi r_1}=\displaystyle\frac{\mu_0	imes Ir_1}{2\pi a^2} Now, magnetic filed at point (B_2)=\displaystyle\frac{\mu_0}{2\pi}\cdot \frac{I}{2a}=\frac{\mu_0I}{4\pi a} 	herefore  Required Ratio  =\displaystyle\frac{B_1}{B_2}=\frac{\mu_0 I r_1}{2\pi a^2}	imes \displaystyle\frac{4\pi a}{\mu_0I} =\displaystyle\frac{2r_1}{a}=\frac{2	imes \displaystyle\frac{a}{2}}{a}=1

Question 3

A long straight wire of radius R carries a steady current  I_o , uniformly distributed throughout the cross-section of the wire. The magnetic field at a radial distance r from the centre of the wire, in the region  r > R , is?

Accepted Answer

Radius of wire = RSteady current passing through the wire =I_o The magnetic field at radial distance r from the centre of the wire \displaystyle\oint B\cdot dl=\mu_oi \Rightarrow B\displaystyle\int dl=\mu_oI_o \Rightarrow B=\dfrac{\mu_oI_o}{2\pi r} .

Question 4

A galvanometer has a coil of resistance  100  Omhs and gives a full-scale deflection for  30  mA current. If it is to work as a voltmeter of  30 V  range, the resistance required to be added will be

Accepted Answer

Given data:-Resistance of Galvanometer =R_{G}=100\Omega Current through galvanometer for full scale deflection =I_{g}=30mA Now if we want to convert galvanometer into voltmeter, then we have to connect a large resistance R in series with galvanometer.Now equivalent Resistance of galvanometer and large resistance R =R_{G}+R Current through galvanometer and large resistance =I_{g}=\dfrac{V}{R_{G}+R}    V=voltage 30*10^{-3}=\dfrac{30}{100+R} \dfrac{30}{1000}=\dfrac{30}{100+R} 100+R=1000 R=900\Omega

Question 5

An  80\Omega  galvanometer deflects full-scale for a potential of 20 mV. A voltmeter deflecting full-scale of 5 V is to be made using this galvanometer. We must connect

Accepted Answer

The current through the galvanometer producing full-scale deflection is I=\dfrac {V}{R} where, variables have their usual meanings I=\dfrac {20	imes 10^{-3}}{80} 	herefore I=2.5	imes 10^{-4}A To convert the galvanometer into a voltmeter, a high resistance is connected in series with the galvanometer. Therefore, R=\dfrac{V}{I_G}-G R =\dfrac{5}{2.5	imes 10^{-4}}-80   R =20000 - 80 R=19920     \Omega R = 19.92   k \Omega

Question 6

A galvanometer coil has a resistance of  150  gives full scale deflection for a current of  4 mA . To convert it to an ammeter of range  0  to  6A

Accepted Answer

Galvanometer resistance  R_g = 150 Current flowing through  R_g ,  I_g = 4 mA = 0.004 A Let a shunt resistance  R_s  is connected in parallel to  R_g  to convert it into an ammeter.Using  (I - I_g) R_s = I_g R_g 	herefore   (6 - 0.004) R_s = 0.004 	imes 150 Or  6 	imes R_s = 0.6 \implies   R_s = 0.1 Hence  0.1  resistance must be connected in parallel to the galvanometer so as to convert it into an ammeter.

Question 7

A galvanometer of 100 resistance gives full scale deflection with 0.01A current. How much resistance should be connected to convert it into an ammeter of range 10A?

Accepted Answer

Given: A galvanometer of  100  resistance gives full scale deflection with  0.01A  currentTo find the resistance that should be connected convert it into an ammeter of range  10A Solution:Galvanometer is a very sensitive instrument therefore it cannot measure heavy currents. In order to convert a Galvanometer into an Ammeter, a very low shunt resistance is connected in parallel to Galvanometer. Value of shunt is so adjusted that most of the current passes through the shunt.If resistance of galvanometer is  R_g  and it gives full-scale deflection when current  I_g  is passed through it. Then, V = I_gR_g\\implies V=0.01A	imes 100\Omega\\implies V=1V Let a shunt of resistance ( R_s ) is connected in parallel to galvanometer. If total current through the circuit is  I . I = 10 A = I_s + I_g V = I_sR_s = (I - I_g)R_s (I - I_g)R_s = I_gR_g (10 - 0.01)R_s = 100 	imes0.01\\implies R_s=0.1\Omega is the resistance that should be connected in parallel to convert it into an ammeter of range  10A

Moving Charges And Magnetism

Tip 3: Galvanometer to Voltmeter

Tip 4: Galvanometer to Ammeter