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<\/figure>\n\n\n\nQuestion 5:<\/h4>\n\n\n\n
Sketch the magnetic field lines for a current-carrying circular loop near its centre. Replace the loop by an equivalent magnetic dipole and sketch the magnetic field lines near the centre of the dipole. Identify the difference.<\/p>\n\n\n\n
Answer:<\/h4>\n\n\n\n
The difference between the two configurations is that in the current-carrying loop, the magnetic field lines pass through the centre and are perpendicular to its axis; whereas in the equivalent magnetic dipole, the magnetic field lines do not pass through the centre.<\/p>\n\n\n\n The force on a north pole,<\/p>\n\n\n\n F\u2192=mB\u2192, parallel to the field<\/p>\n\n\n\n B\u2192. Does it contradict our earlier knowledge that a magnetic field can exert forces only perpendicular to itself?<\/p>\n\n\n\n Yes, it seems to contradict with our earlier knowledge that a magnetic field can exert forces only perpendicular to itself.<\/p>\n\n\n\n F\u2192=mB\u2192Here,<\/p>\n\n\n\n B\u2192= Magnetic field Two bar magnets are placed close to each other with their opposite poles facing each other. In absence of other forces, the magnets are pulled towards each other and their kinetic energy increases. Does it contradict our earlier knowledge that magnetic forces cannot do any work and hence cannot increase kinetic energy of a system?<\/p>\n\n\n\n Yes, it contradicts our earlier knowledge that magnetic forces cannot do any work and hence cannot increase the kinetic energy of the system. When opposite poles are facing each other, an attractive force acts between them so the magnets are pulled towards each other. As the two magnets come close to each other so the force between them increases and hence, the kinetic energy also increases.<\/p>\n\n\n\n Magnetic scalar potential is defined as<\/p>\n\n\n\n Ur\u21922-Ur\u21921=-\u222br\u21921r\u21922B\u2192.dl\u2192Apply this equation to a closed curve enclosing a long straight wire. The RHS of the above equation is then \u2212\u03bc0<\/sub> i<\/em> by Ampere\u2019s law. We see that<\/p>\n\n\n\n Ur\u21922\u2260Ur\u21921even when<\/p>\n\n\n\n r\u21922=r\u21921. Can we have a magnetic scalar potential in this case?<\/p>\n\n\n\n No, we cannot have a magnetic scalar potential here.<\/p>\n\n\n\n Ampere\u2019s law is a method of calculating magnetic field due to current distribution. On the other hand, magnetic scalar potential requires a magnetic field due to pole strength m<\/em>. Can the earth\u2019s magnetic field be vertical at a place? What will happen to a freely suspended magnet at such a place? What is the value of dip here?<\/p>\n\n\n\n Yes, Earth\u2019s magnetic field is vertical at the poles. A freely suspended magnet becomes vertical at the poles, with its north pole pointing towards Earth\u2019s north pole, which is magnetic south. Can the dip at a place be (a) zero (b) 90\u00b0?<\/p>\n\n\n\n (a) Yes, the dip can be zero at the equator of Earth.<\/p>\n\n\n\n (b) Yes, the dip can be 90\u00b0\u00e2\u20ac\u2039 at the poles of Earth.<\/p>\n\n\n\n The reduction factor K of a tangent galvanometer is written on the instrument. The manual says that the current is obtained by multiplying this factor to tan \u03b8. The procedure works well at Bhuwaneshwar. Will the procedure work if the instrument is taken to Nepal? If there is same error, can it be corrected by correcting the manual or the instrument will have to be taken back to the factory?<\/p>\n\n\n\n Yes, the procedure will work if the instrument is taken to Nepal, as the current at a place can be calculated by multiplying the reduction factor K with tan<\/p>\n\n\n\n \u03b8 of that place. In our case, we will take the value of tan<\/p>\n\n\n\n \u03b8 of Nepal, as tan<\/p>\n\n\n\n \u03b8may vary from place to place. tan<\/p>\n\n\n\n \u03b8at any place is determined from the mathematical formula<\/p>\n\n\n\n BBH, where B<\/em> is the external magnetic field and BH<\/sub> <\/em>is the horizontal component of Earth\u2019s magnetic field. Thus, we need not take the manual or the instrument back to the factory for correction.<\/p>\n\n\n\n A circular loop carrying a current is replaced by an equivalent magnetic dipole. A point on the axis of the loop is in<\/p>\n\n\n\n (a) end-on position (a) end-on position<\/p>\n\n\n\n Points lying on the axis of a magnet are called end-on points. In our case, the point on the axis of the loop (on replacing the circular loop with an equivalent magnetic dipole) lies on the axis of the magnetic dipole or on the end-on position. A circular loop carrying a current is replaced by an equivalent magnetic dipole. A point on the loop is in<\/p>\n\n\n\n (a) end-on position (b) broadside-on position<\/p>\n\n\n\n The position of the points lying on the equator of a magnetic dipole is called the broadside-on position. In our case, the point on the loop (after replacement of the circular loop with an equivalent magnetic dipole) lies on the equatorial position of the equivalent magnetic dipole. Hence, the point lies on the broadside-on position. When a current in a circular loop is equivalently replaced by a magnetic dipole,<\/p>\n\n\n\n (a) the pole strength m<\/em> of each pole is fixed (c) the product md<\/em> is fixed<\/p>\n\n\n\n When we replace a circular current-carrying loop with a magnetic dipole to resemble field lines of the circular loop, the pole strength m <\/em>and the distance between the poles are not fixed. Let r<\/em> be the distance of a point on the axis of a bar magnet from its centre. The magnetic field at such a point is proportional to<\/p>\n\n\n\n (a)<\/p>\n\n\n\n 1r(b)<\/p>\n\n\n\n 1r2(c)<\/p>\n\n\n\n 1r3(d) none of these<\/p>\n\n\n\n (d) None of these<\/p>\n\n\n\n Magnetic field B<\/em> due to a bar magnet of magnetic moment M<\/em> at distance r <\/em>of the point on the axis of the magnet from its centre is given by<\/p>\n\n\n\n B=\u03bco4\u03c02Mrr2-l22 B\u221drr2-l22.<\/p>\n\n\n\n Let r<\/em> be the distance of a point on the axis of a magnetic dipole from its centre. The magnetic field at such a point is proportional to<\/p>\n\n\n\n (a)<\/p>\n\n\n\n 1r(b)<\/p>\n\n\n\n 1r2(c)<\/p>\n\n\n\n 1r3(d) none of these<\/p>\n\n\n\n (c)<\/p>\n\n\n\n 1r3Magnetic field B<\/em> due to a bar magnet of magnetic moment M<\/em> at distance r<\/em> of the point on the axis from its centre is given by<\/p>\n\n\n\n B=\u03bc02Mr4\u03c0r2-l22Here, 2l <\/em>is the length of the magnet. r >>l, the bar magnet acts like a magnetic dipole whose magnetic field is \u221d 1r3Now, l<\/em> in the denominator can be neglected. Two short magnets of equal dipole moments M are fastened perpendicularly at their centre (figure 36-Q1). The magnitude of the magnetic field at a distance d<\/em> from the centre on the bisector of the right angle is<\/p>\n\n\n\n (a)<\/p>\n\n\n\n \u03bc04\u03c0Md3(b)<\/p>\n\n\n\n \u03bc04\u03c02 Md3(c)<\/p>\n\n\n\n \u03bc04\u03c022Md3(d)<\/p>\n\n\n\n \u03bc04\u03c02Md3Figure<\/p>\n\n\n\n (c)<\/p>\n\n\n\n \u03bc04\u03c022Md3 B\u21921=\u03bc04\u03c02Md3 \u20261Magnetic field (B<\/em>2<\/sub>) due to the short dipole B of dipole moment M<\/em> at an axial point is given by,<\/p>\n\n\n\n B\u21922=\u03bc04\u03c02Md3 \u20262Resultant magnetic field (B<\/em>) will be, B12+B22B<\/em> =<\/p>\n\n\n\n \u03bc04\u03c022Md3<\/p>\n\n\n\n Magnetic meridian is<\/p>\n\n\n\n (a) a point (d) a vertical plane<\/p>\n\n\n\n Magnetic meridian at a place is not a line but a vertical plane passing through the axis of a freely suspended magnet.<\/p>\n\n\n\n A compass needle which is allowed to move in a horizontal plane is taken to a geomagnetic pole. It<\/p>\n\n\n\n (a) will stay in north-south direction only (d) will stay in any position<\/p>\n\n\n\n When taken to a geomagnetic pole, a compass needle that is allowed to move in a horizontal plane will try to suspend itself vertically to the horizontal plane containing the compass. In other words, the horizontal plane containing the compass will restrict the compass to suspend itself in vertical direction; hence, the compass will stay in any position. A dip circle is taken to geomagnetic equator. The needle is allowed to move in a vertical plane perpendicular to the magnetic meridian. The needle will stay<\/p>\n\n\n\n (a) in horizontal direction only (d) in the direction it is released<\/p>\n\n\n\n At the geomagnetic equator, the needle tries to suspend itself in horizontal direction. But here the needle is restricted to move only in the vertical plane perpendicular to the magnetic meridian. Hence, the needle will stay in the direction it is released.<\/p>\n\n\n\n Which of the following four graphs may best represent the current-deflection relation in a tangent galvanometer?<\/p>\n\n\n\n Figure<\/p>\n\n\n\n (c) curve<\/p>\n\n\n\n Since i<\/em><\/p>\n\n\n\n \u221dtan<\/p>\n\n\n\n \u03b8, the only graph that represents this correlation is curve c<\/em>.<\/p>\n\n\n\n A tangent galvanometer is connected directly to an ideal battery. If the number of turns in the coil is doubled the deflection will<\/p>\n\n\n\n (a) increase (c) remain unchanged<\/p>\n\n\n\n For a tangent galvanometer, deflection is given by<\/p>\n\n\n\n \u03b8 = tan-1ik If the current is doubled, the deflection is also doubled in<\/p>\n\n\n\n (a) a tangent galvanometer (b) a moving coil galvanometer<\/p>\n\n\n\n The current and deflection dependence of a moving coil galvanometer is given by<\/p>\n\n\n\n i=knAB\u03b8 \u21d2i\u221d\u03b8Therefore, if we double the current, the deflection also gets doubled. i\u221dtan\u03b8; that is, there is no direct relation between<\/p>\n\n\n\n \u03b8and current. A very long bar magnet is placed with its north pole coinciding with the centre of a circular loop carrying as electric current i<\/em>. The magnetic field due to the magnet at a point on the periphery of the wire is B. The radius of the loop is a<\/em>. The force on the wire is<\/p>\n\n\n\n (a) very nearly 2\u03c0ai<\/em>B perpendicular to the plane of the wire (a) very nearly 2<\/p>\n\n\n\n \u03c0aiB perpendicular to the plane of the wire<\/p>\n\n\n\n In this case, the north pole of the magnet is coinciding with the centre of the circular loop carrying electric current i.<\/em> So, the magnetic field lines almost lie on the plane of the ring and the force due to the field lines is perpendicular to the field lines and to the plane of the circular ring. \u21d2F=\u222b02\u03c0aBidl<\/p>\n\n\n\n \u21d2F=2\u03c0aiBThus, the force acting on the wire is 2<\/p>\n\n\n\n \u03c0ai<\/em>B and it is perpendicular to the plane of the wire.<\/p>\n\n\n\n Pick the correct options.<\/p>\n\n\n\n (a) Magnetic field is produced by electric charges only (a) Magnetic field is produced by electric charges only. Justification of (a) and (b):<\/p>\n\n\n\n Investigators and experimenters have failed to find any sign of magnetic monopoles. So, we can assume that magnetic monopoles are only a mathematical assumption. Denial of (c):<\/p>\n\n\n\n The north pole is equivalent to an anticlockwise current and the south pole is equivalent to a clockwise current.<\/p>\n\n\n\n Denial of (d):<\/p>\n\n\n\n A bar magnet is not equivalent to a long, straight current because the distribution and orientation of magnetic field lines do not resemble each other.<\/p>\n\n\n\n A horizontal circular loop carries a current that looks clockwise when viewed from above. It is replaced by an equivalent magnetic dipole consisting of a south pole S and a north pole N.<\/p>\n\n\n\n (a) The line SN should be along a diameter of the loop. (b) The line SN should be perpendicular to the plane of the loop A horizontal circular loop carrying current in clockwise direction acts like the south pole of a magnet. Hence, the south pole of the magnet coincides with the loop. Now, when the loop carrying current in clockwise direction is viewed from above, it looks like the magnetic lines of force are entering the loop thus it acts like south pole of a magnet. And if we view from below the loop then it appears that magnetic lines of force are leaving the loop. Hence, the north pole should be below the loop.<\/p>\n\n\n\n Consider a magnetic dipole kept in the north to south direction. Let P1<\/sub>, P2<\/sub>, Q1<\/sub>, Q2<\/sub> be four points at the same distance from the dipole towards north, south, east and west of the dipole respectively. The directions of the magnetic field due to the dipole are the same at<\/p>\n\n\n\n (a) P1<\/sub> and P2<\/sub> (a) P1<\/sub> and P2<\/sub> We know that magnetic field lines are directed from the north pole to the south pole. From the given figure, we can say that the direction of magnetic field<\/p>\n\n\n\n \u2192Bis the same only at points P1<\/sub> and P2<\/sub> and at points Q1<\/sub> and Q2<\/sub>.<\/p>\n\n\n\n Consider the situation of the previous problem. The directions of the magnetic field due to the dipole are opposite at<\/p>\n\n\n\n (a) P1<\/sub> and P2<\/sub> (c) P1<\/sub> and Q1<\/sub> We know that magnetic field lines are directed from the north pole to the south pole. From the given figure, we can say that the direction of the magnetic field<\/p>\n\n\n\n \u2192Bis opposite at points P1<\/sub> and Q1<\/sub> and at points P2<\/sub> and Q2<\/sub>.<\/p>\n\n\n\n To measure the magnetic moment of a bar magnet, one may use<\/p>\n\n\n\n (a) a tangent galvanometer (b) a deflection galvanometer if the earth\u2019s horizontal field is known Denial of (a):<\/p>\n\n\n\n Tangent galvanometer is an instrument used to measure electric current; it cannot be used to the measure magnetic moment of a bar magnet.<\/p>\n\n\n\n Justification of (b) and (c):<\/p>\n\n\n\n Deflection magnetometer is used to measure<\/p>\n\n\n\n MBH of a permanent bar magnet. Justification of (d):<\/p>\n\n\n\n Using deflection and oscillation magnetometers, we can calculate<\/p>\n\n\n\n MBHand M BH<\/sub><\/em>, respectively. Therefore, if we multiply the result obtained from both the instruments, then BH<\/sub> <\/em>cancels out as<\/p>\n\n\n\n MBH \u00d7 M BH<\/sub><\/em>\u00e2\u20ac\u2039 = M2<\/sup>.<\/em> Thus, the value of BH<\/sub><\/em> is not required. A long bar magnet has a pole strength of 10 Am. Find the magnetic field at a point on the axis of the magnet at a distance of 5 cm from the north pole of the magnet.<\/p>\n\n\n\n Given: Bis given by<\/p>\n\n\n\n B=\u03bc04\u03c0 mr2 =10-7\u00d7105\u00d710-22=10-625\u00d710-4 =10-225=4\u00d710-4 T<\/p>\n\n\n\n Two long bare magnets are placed with their axes coinciding in such a way that the north pole of the first magnet is 2.0 cm from the south pole of the second. If both the magnets have a pole strength of 10 Am, find the force exerted by one magnet of the other.<\/p>\n\n\n\n Given: Fexerted by two magnetic poles on each other is given by<\/p>\n\n\n\n F=\u03bc04\u03c0m1m2r2 =4\u03c0\u00d710-7\u00d71024\u03c0\u00d74\u00d710-4 =2.5\u00d710-2 N<\/p>\n\n\n\n A uniform magnetic field of 0.20 \u00d7 10\u22123<\/sup> T exists in the space. Find the change in the magnetic scalar potential as one moves through 50 cm along the field.<\/p>\n\n\n\n Given: \u2206r= 50 cm B =-dVdl \u21d2dV = \u222br1r2 -B.\u21c0dl \u21d2 \u2206V = -B.(\u2206r)Here,<\/p>\n\n\n\n \u2206Vis the change in the potential. Figure (36-E1) shows some of the equipotential surfaces of the magnetic scalar potential. Find the magnetic field B at a point in the region.<\/p>\n\n\n\n Figure<\/p>\n\n\n\n Given: B=-dVdx\u21d2B = -0.1\u00d710-4 T-m5\u00d710-2 m \u21d2B = -2\u00d710-4 TB<\/em> is perpendicular to the equipotential surface. Here, it is at angle of 120\u00b0 with the positive x<\/em>-axis.<\/p>\n\n\n\n The magnetic field at a point, 10 cm away from a magnetic dipole, is found to be 2.0 \u00d7 10\u22124<\/sup> T. Find the magnetic moment of the dipole if the point is (a) in end-on position of the dipole and (b) in broadside-on position of the dipole.<\/p>\n\n\n\n Given: (a) If the point is at the end-on position: Bon the axial point of the dipole is given by<\/p>\n\n\n\n B=\u03bc04\u03c02Md3Here, M is the magnetic moment of the dipole that we need to find out.\u22342\u00d710-4=10-7\u00d72M10-13\u21d2M=2\u00d710-4\u00d710-310-7\u00d72\u21d2M=1 A-m2(b) If the point is at broadside-on position (equatorial position): Bis given by<\/p>\n\n\n\n B=\u03bc04\u03c0Md3\u21d22\u00d710-4=10-7\u00d7M10-13\u21d2M=2 A-m2<\/p>\n\n\n\n Show that the magnetic field at a point due to a magnetic dipole is perpendicular to the magnetic axis if the line joining the point with the centre of the dipole makes an angle of<\/p>\n\n\n\n tan-1 2with the magnetic axis.<\/p>\n\n\n\n Given: \u03b8=<\/p>\n\n\n\n tan-12<\/p>\n\n\n\n \u21d2tan \u03b8=2 \u21d22=tan2 \u03b8\u21d2tan \u03b8= cot \u03b8\u21d2tan \u03b82=cot \u03b8 \u2026.1We know,tan \u03b82=tan \u03b1 \u2026.2 \u21d2tan \u03b1 = tan (90 \u2212 \u03b8<\/em>)<\/p>\n\n\n\n \u21d2\u03b1 = 90 \u2212 \u03b8<\/em><\/p>\n\n\n\n \u21d2\u03b8<\/em> + \u03b1 = 90\u00b0 A bar magnet has a length of 8 cm. The magnetic field at a point at a distance 3 cm from the centre in the broadside-on position is found to be 4 \u00d7 10\u22126<\/sup> T. Find the pole strength of the magnet.<\/p>\n\n\n\n Given: \u00d710<\/p>\n\n\n\n -2m Bis given by<\/p>\n\n\n\n B=\u03bc0m2l4\u03c0d2+l23\/2Here, m<\/em> is the pole strength of the magnet. 4\u00d710-6 = 10-7m\u00d78\u00d710-29\u00d710-4+16\u00d710-43\/2\u21d2 4\u00d710-6 = m\u00d78\u00d710-9253\/2\u00d710-43\/2\u21d2m = 4\u00d710-6\u00d7125\u00d710-68\u00d710-9\u21d2m= 6.25\u00d710-2 A-mThus, the pole strength of the magnet is 6.25<\/p>\n\n\n\n \u00d710<\/p>\n\n\n\n -2A-m.<\/p>\n\n\n\n A magnetic dipole of magnetic moment 1.44 A m2<\/sup> is placed horizontally with the north pole pointing towards north. Find the position of the neutral point if the horizontal component of the earth\u2019s magnetic field is 18 \u03bcT.<\/p>\n\n\n\n Given: B\u2192 = \u03bc0M4\u03c0d3This magnetic field strength should be equal to the horizontal component of Earth\u2019s magnetic field at that point, that is, B<\/em>H<\/sub>due to Earth. B\u2192 = \u03bc0M4\u03c0d3\u21d2B=10-7\u00d71.44d3\u21d218\u00d710-6=10-7\u00d71.44d3\u21d2d3=8\u00d710-3\u21d2d=2\u00d710-1=20 cm<\/p>\n\n\n\n A magnetic dipole of magnetic moment 0.72 A m2<\/sup> is placed horizontally with the north pole pointing towards south. Find the position of the neutral point if the horizontal component of the earth\u2019s magnetic field is 18 \u03bcT.<\/p>\n\n\n\n Given: B=\u03bc04\u03c02Md3This magnetic field strength should be equal to the horizontal component of Earth\u2019s magnetic field. 10-7\u00d72\u00d70.72d3=18\u00d710-6\u21d2d3=2\u00d70.72\u00d710-718\u00d710-6\u21d2d= 8\u00d710-910-61\/3\u21d2d=2\u00d710-1 m=20 cm<\/p>\n\n\n\n A magnetic dipole of magnetic moment<\/p>\n\n\n\n 0.722 A m2is placed horizontally with the north pole pointing towards east. Find the position of the neutral point if the horizontal component of the earth\u2019s magnetic field is 18 \u03bcT.<\/p>\n\n\n\n Given: =0.722 A-m2Horizontal component of the earth\u2019s magnetic field, B<\/em>H<\/sub> = 18 \u03bcT Bon the equatorial line of the dipole is given by,<\/p>\n\n\n\n B=\u03bc04\u03c0 Md3\u21d24\u03c0\u00d710-74\u03c0\u00d70.722d3=18\u00d710-6\u21d2d3=0.72\u00d71.414\u00d710-718\u00d710-6\u21d2d3=0.005656\u21d2d=0.0056563\u21d2d\u22480.2 m=20 cm<\/p>\n\n\n\n The magnetic moment of the assumed dipole at the earth\u2019s centre is 8.0 \u00d7 1022<\/sup> A m2<\/sup>. Calculate the magnetic field B at the geomagnetic poles of the earth. Radius of the earth is 6400 km.<\/p>\n\n\n\n Given: B is given by<\/p>\n\n\n\n B=\u03bc04\u03c02Md3\u21d2B= 10-7\u00d72\u00d78\u00d71022643\u00d71015\u21d2B= 6\u00d710-5 T=60 \u03bcT<\/p>\n\n\n\n If the earth\u2019s magnetic field has a magnitude 3.4 \u00d7 10\u22125<\/sup> T at the magnetic equator of the earth, what would be its value at the earth\u2019s geomagnetic poles?<\/p>\n\n\n\n Given: B is given by<\/p>\n\n\n\n B=\u03bc04\u03c0MR3\u21d2\u03bc04\u03c0\u00d7MR3=3.4\u00d710-5\u21d2M=3.4\u00d710-5\u00d7R3\u00d74\u03c04\u03c0\u00d710-7\u21d2M=3.4\u00d7102 R3As the magnetic field on Earth\u2019s geomagnetic poles lies on the axial point of the magnetic dipole, the magnetic field at the axial point<\/p>\n\n\n\n B1is given by<\/p>\n\n\n\n B\u21921=\u03bc04\u03c0\u00d72MR3\u21d2B1=10-7\u00d72\u00d73.4\u00d7102 \u00d7R3R3\u21d2B1=6.8\u00d710-5 T<\/p>\n\n\n\n The magnetic field due to the earth has a horizontal component of 26 \u03bcT at a place where the dip is 60\u00b0. Find the vertical component and the magnitude of the field.<\/p>\n\n\n\n Given: BH is given by![\"\"\/](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/5_Short-answer.png\")
<\/figure>\n\n\n\n![\"\"\/](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/5q11.png\")
<\/figure>\n\n\n\nQuestion 6:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
m<\/em> = Magnetic charge
For a positive magnetic charge, force is along the magnetic field.
For a negative magnetic charge, force is opposite to the magnetic field.
Thus, it contradicts the notion that a magnetic field can exert forces only perpendicular to itself.F<\/p>\n\n\n\nQuestion 7:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 8:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Potential at a distance r<\/em> is given by
\u03bc0m4\u03c0rAs there is no current distribution, no magnetic field due to poles or the pole strength is present. That is why we cannot have a magnetic scalar potential in this case.<\/p>\n\n\n\nQuestion 9:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
The value of the angle of the dip here is 90\u00b0.<\/p>\n\n\n\nPage No 276:<\/h4>\n\n\n\n
Question 10:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 11:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 1:<\/h4>\n\n\n\n
(b) broadside-on position
(c) both
(d) none<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
If P was the point on the axis of the loop, then it is clear from the figure that P lies on the end-on position of the equivalent magnetic dipole.<\/p>\n\n\n\n![\"\"\/](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/6q75.png\")
<\/figure>\n\n\n\nQuestion 2:<\/h4>\n\n\n\n
(b) broadside-on position
(c) both
(d) none<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
If P was the point on the loop, then it is clear from the figure that point P lies on the broadside-on position of the equivalent magnetic dipole.<\/p>\n\n\n\n![\"\"\/](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/7q72.png\")
<\/figure>\n\n\n\nQuestion 3:<\/h4>\n\n\n\n
(b) the distance d<\/em> between the poles is fixed
(c) the product md <\/em>is fixed
(d) none of the above<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
But the magnetic dipole moment of both systems is always fixed. It is the product of the magnetic moment and the distance between the poles. In other words, md <\/em>is fixed.
A current loop of area A<\/em> and current I<\/em> can be replaced with a magnetic dipole of dipole moment md.<\/em>
i.e. md <\/em>= IA<\/em><\/p>\n\n\n\nQuestion 4:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Here, 2l <\/em>is the length of the magnet.
So, from the above formula, it can be easily seen that<\/p>\n\n\n\nQuestion 5:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
When the distance of the point where the magnetic field has to be calculated is greater than the length of the magnet, i.e<\/p>\n\n\n\n
B<\/em><\/p>\n\n\n\n
So, the correct option is (c).<\/p>\n\n\n\nQuestion 6:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/Q619.png\")
Magnetic field (B<\/em>1<\/sub>) due to the short dipole A of dipole moment M<\/em> at an axial point is given by,<\/p>\n\n\n\n
B<\/em> =<\/p>\n\n\n\nQuestion 7:<\/h4>\n\n\n\n
(b) a line along north-south
(c) a horizontal plane
(d) a vertical plane<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 8:<\/h4>\n\n\n\n
(b) will stay in east-west direction only
(c) will become rigid showing no movement
(d) will stay in any position<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
However, a freely suspended magnet will become vertical at poles, with its north pole pointing towards Earth at its north pole (which is magnetic south).<\/p>\n\n\n\nQuestion 9:<\/h4>\n\n\n\n
(b) in vertical direction only
(c) in any direction except vertical and horizontal
(d) in the direction it is released<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 10:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 11:<\/h4>\n\n\n\n
(b) decrease
(c) remain unchanged
(d) either increase or decrease<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Here, k<\/em> is the constant called reduction factor.
From the above formula, we can say that deflection is independent of the number of turns.
Hence, on doubling the number of turns, deflection remains the same.<\/p>\n\n\n\nQuestion 12:<\/h4>\n\n\n\n
(b) a moving-coil galvanometer
(c) both
(d) none<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
However, in a tangent galvanometer,<\/p>\n\n\n\n
Hence, the correct option is (b).<\/p>\n\n\n\nQuestion 13:<\/h4>\n\n\n\n
(b) 2\u03c0ai<\/em>B in the plane of the wire
(c) \u03c0ai<\/em>B along the magnet
(d) zero<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
![\"\"\/](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/13172.png\")
<\/figure>\n\n\n\n
Let idl <\/em>be the current element, B be the magnetic field and dF <\/em>be the force on the current element idl.<\/em>
Now
dF = <\/em>Bidl<\/em><\/p>\n\n\n\nPage No 277:<\/h4>\n\n\n\n
Question 1:<\/h4>\n\n\n\n
(b) Magnetic poles are only mathematical assumptions having no real existence
(b) A north pole is equivalent to a clockwise current and a south pole is equivalent to an anticlockwise current.
(d) A bar magnet is equivalent to a long, straight current.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
(b) Magnetic poles are only mathematical assumptions having no real existence.<\/p>\n\n\n\n
A magnetic field is produced by the motion of an electric charge only. In paramagnets or ferromagnets, the motion of an electron (charge) and the alignment of domains (bunch of charges with particular alignment) create paramagnetism and ferromagnetism, respectively.
Therefore, the only cause behind the magnetic field is the motion of an electric charge.<\/p>\n\n\n\nQuestion 2:<\/h4>\n\n\n\n
(b) The line SN should be perpendicular to the plane of the loop
(c) The south pole should be slow the loop
(d) The north pole should be below the loop<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
(d) The north pole should be below the loop.<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/2830.png\")
<\/p>\n\n\n\nQuestion 3:<\/h4>\n\n\n\n
(b) Q1<\/sub> and Q2<\/sub>
(c) P1<\/sub> and Q1<\/sub>
(d) P2<\/sub> and Q2<\/sub><\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
(b) Q1<\/sub> and Q2<\/sub><\/p>\n\n\n\n![\"\"\/](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/3665.png\")
<\/figure>\n\n\n\nQuestion 4:<\/h4>\n\n\n\n
(b) Q1<\/sub> and Q2<\/sub>
(c) P1<\/sub> and Q1<\/sub>
(d) P2<\/sub> and Q2<\/sub><\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
(d) P2<\/sub> and Q2<\/sub><\/p>\n\n\n\n![\"\"\/](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/3665-1.png\")
<\/figure>\n\n\n\nQuestion 5:<\/h4>\n\n\n\n
(b) a deflection galvanometer if the earth\u2019s horizontal field is known
(c) an oscillation magnetometer if the earth\u2019s horizontal field is known
(d) both deflection and oscillation magnetometer if the earth\u2019s horizontal field is not known<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
(c) an oscillation magnetometer if the earth\u2019s horizontal field is known
(d) both deflection and oscillation magnetometers if the earth\u2019s horizontal field is not known<\/p>\n\n\n\n
Similarly, oscillation magnetometer is used to measure M BH<\/sub> <\/em>of a bar magnet. So, if earth\u2019s horizontal field, BH<\/sub><\/em>, is known, then the magnetic moment of a bar magnet, M, <\/em>can be measured.<\/p>\n\n\n\n
Therefore, we can use both deflection and oscillation magnetometers if the earth\u2019s horizontal field is not known.<\/p>\n\n\n\nQuestion 1:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Pole strength of the bar magnet, m<\/em> = 10 Am
Distance of the point from the north pole of the bar magnet, r <\/em>= 5 cm = 0.05 m
We know,
The magnetic field due to magnetic charge<\/p>\n\n\n\nQuestion 2:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Pole strength = m<\/em>1<\/sub> = m<\/em>2<\/sub> = 10 Am
Distance between the north pole of the first magnet and the south pole of the second magnet, r<\/em> = 2 cm = 0.02 m
We know,
Force<\/p>\n\n\n\nQuestion 3:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Magnetic field in the space, B<\/em> = 0.20 \u00d7 10\u22123<\/sup> T
Distance moved,<\/p>\n\n\n\n
We know,<\/p>\n\n\n\n
Since the magnetic field is uniform, it can come out of the integration sign.<\/p>\n\n\n\n
\u2234 Change in the potential = \u22120.2 \u00d7 10\u22123<\/sup> \u00d7 0.5
= \u22120.1 \u00d7 10\u22123<\/sup> T-m
Here, the negative sign shows that the potential decreases.<\/p>\n\n\n\nQuestion 4:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Perpendicular distance, dx<\/em> = 10 sin30\u00e2\u201a\u2019<\/sup> cm = 0.05 m,
Change in the potential, dV<\/em> = 0.1 \u00d7 10\u22124<\/sup> T-m
We know that the relation between the potential and the field is given by<\/p>\n\n\n\nQuestion 5:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Magnetic field strength, B<\/em> = 2 \u00d7 10\u22124<\/sup> T
Distance of the point from the dipole, d<\/em> = 10 cm = 0.1 m<\/p>\n\n\n\n
The magnetic field<\/p>\n\n\n\n
The magnetic field<\/p>\n\n\n\nQuestion 6:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Angle made by observation point P with the axis of the dipole,<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/Q6_pg-2771.png\")
On comparing (1) and (2), we get
tan \u03b1 = cot \u03b8<\/em><\/p>\n\n\n\n
Hence, the magnetic field due to the dipole is perpendicular to the magnetic axis.<\/p>\n\n\n\nQuestion 7:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Length of the magnet, 2l<\/em> = 8 cm = 8<\/p>\n\n\n\n
Distance of the observation point from the centre of the dipole, d<\/em> = 3 cm
Magnetic field in the broadside-on position, B<\/em> = 4 \u00d7 10\u22126<\/sup> T
The magnetic field due to the dipole on the equatorial point<\/p>\n\n\n\n
On substituting the respective values, we get<\/p>\n\n\n\nQuestion 8:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Magnetic moment of the magnetic dipole, M = <\/em>1.44 A-m2<\/sup>
\u00e2\u20ac\u2039Horizontal component of Earth\u2019s magnetic field, B<\/em>H<\/sub>= 18 \u03bcT
We know that for a magnetic dipole with pole pointing to the north, the neutral point always lies in the broadside-on position.
Let d<\/em> be the perpendicular distance of the neutral point from mid point of the magnet.
The magnetic field due to the dipole at the broadside-on position (B<\/em>) is given by<\/p>\n\n\n\n
Thus,<\/p>\n\n\n\nQuestion 9:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Magnetic moment of the magnetic dipole, M = <\/em>0.72 Am2<\/sup>
\u00e2\u20ac\u2039Horizontal component of Earth\u2019s magnetic field, B<\/em>H<\/sub>\u00e2\u20ac\u2039 = 18 \u03bcT
\u00e2\u20ac\u2039Let d<\/em> be the distance of the neutral point from the south of the dipole.
When the magnet is such that its north pole faces the geographic south of Earth, the neutral point lies along the axial line of the magnet.
Thus, the magnetic field on the axial point of the dipole (B<\/em>) is given by<\/p>\n\n\n\n
Thus,<\/p>\n\n\n\nQuestion 10:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Magnetic moment of magnetic dipole, M<\/em><\/p>\n\n\n\n
Let d<\/em> be the distance of neutral point from the dipole.
Magnetic field due to the bar magnet<\/p>\n\n\n\nQuestion 11:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Magnetic moment of dipole at Earth\u2019s centre, M<\/em> = 8.0 \u00d7 1022<\/sup> Am2<\/sup>
Radius of Earth, d<\/em> = 6400 km
The geomagnetic pole is at the end of the position (axial) of Earth.
Thus, the magnetic field at the axial point of the dipole<\/p>\n\n\n\nQuestion 12:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Magnetic field at the magnetic equator, B = <\/em>3.4 \u00d7 10\u22125<\/sup> T
Let M<\/em> be the magnetic moment of Earth\u2019s magnetic dipole and R be the distance of the observation point from the centre of Earth\u2019s magnetic dipole.
As the point on the magnetic equator is on the equatorial position of Earth\u2019s magnet, the magnetic field at the equatorial point<\/p>\n\n\n\nQuestion 13:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Horizontal component of Earth\u2019s magnetic field, B = <\/em>26 \u03bcT \u00e2\u20ac\u2039
Angle of dip, \u03b4 = 60\u00b0
The horizontal component of Earth\u2019s magnetic field<\/p>\n\n\n\n
BH<\/sub><\/em> = B<\/em>cos\u03b4