Here the needle and vertical scale are in a plane perpendicular to the magnetic meridian. The box is then rotated through 90 o position on the horizontal circular scale. Then the angle of dip is A dip circle is taken to geomagnetic equator. For a vertical plane other than magnetic meridian $\alpha>0$ or $\cos \alpha<1$ so $\theta^{\prime}>\theta$ In a vertical plane other than magnetic meridian angle of dip is more than in magnetic meridian. perpendicular to geographic axis of Earth. If BH = 0.314 × 10^-4 T , the current in the coil is : The same needle is then allowed to vibrate in the horizontal plane and the time period is again found to be 2 seconds. Important Solutions 17 ... perpendicular to the magnetic axis of Earth. The same needle is then allowed to vibrate in the horizontal plane and the time period is again found to be 2 sec. Maharashtra State Board HSC Science (Electronics) 11th. Click hereto get an answer to your question ️ A vertical circular coil of radius 0.1 m and having 10 turns carries a steady current. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at … The needle will stay (a) in horizontal direction only (b) in vertical direction only (c) in any direction except vertical and horizontal (d) in the direction it is released The time period of vibration is found to be 2 seconds. A dip needle vibrates in the vertical plane perpendicular to the magnetic meridian. What is the amount of work done by the magnetic field? Reuse & Permissions × A magnetic needle free to rotate about the vertical direction (compass ) point west of the geographic north . Magnetic meridian is the plane . The needle is allowed to move in a vertical plane perpendicular to the magnetic meridian. An electron with charge -e and mass m travels at a speed v in a plane perpendicular to a magnetic field of magnitude B. Textbook Solutions 6926. Real dip theta=40.9^@ Let the vertical component of earth's magnetic field be V and its horizontal component at magnetic meridian be H. Then the angle of dip theta at magnetic meridian will be given by tantheta=V/H.....[1] When the dip needle is suspended at an angle of 30^@ to the earth magnetic meridian then it makes an angle 45^@ with the horizontal. The electron follows a circular path of radius R. In a time, t, the electron travels halfway around the circle. A dip needle vibrates in the vertical plane perpendicular to the magnetic meridian. To get the true heading, you need to first read the magnetic compass, then either add an Easterly, or subtract a Westerly, magnetic variation; based on the isogonic lines. Then the angle of dip is (a) 0° (b) 30° (c) 45° (d) 90° If the magnetic field is increased from 0.01 T to 0.06 T during a time interval of … When the plane of the coil is normal to the magnetic meridian, a neutral point is observed at the centre of the coil. A magnetic field B is applied perpendicular to the plane of the sample. The magnitude of the horizontal component of the earth ' magnetic field at the place is known to be . A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22° with the horizontal. passing through the magnetic axis of Earth. Another magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at with the horizontal . East of this agonic line, the magnetic North Pole is to the west of the geographic North Pole and a correction must be applied to a compass indication to get a true direction. 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