The following M.C.Q. which appeared in the All India Pre –Medical/Dental Entrance-2005 (C.B.S.E) test paper will be interesting to you:
An electron moves in a circular orbit with a uniform speed ‘v’. It produces a magnetic field ‘B’ at the centre of the circle. The radius of the circle is proportional to
(a) √(B/v) (b) B/v (c) √(v/B) (d) v/B
The electron, moving along the circular path is equivalent to a current I = e/T = ev/(2πr) where ‘e’ is the electronic charge, T is the period, ‘v’ is the speed and ‘r’ is the radius of the circular motion. The magnetic field at the centre of the circle, B = μ0I/2r = μ0ev/4πr2 so that r α √(v/B). The correct option is (c).
Now, consider the following simple question. (Be careful however, so that you won’t pick out a wrong answer):
A long straight conductor P carries a current ‘I’ while another parallel long straight conductor Q carries a current 2I in the same direction. The magnetic field midway between them is B. If the current 2I in the conductor Q is switched off, the magnetic field midway between the conductors will be
(a) B/3 (b) B (c) –B (d) –B/2 (e) 2B
The magnetic fields produced by the conductors are in opposition in the region between them. The field produced by Q is twice the field produced by P. If the field produced by Q is 2B, the field produced by P is –B. This is indeed the case, since the net field is B when both currents are switched on. When the current 2I in Q is switched off, the field due to Q alone is present. The answer therefore is –B [Option (c)].
The following question also is simple. It is designed to check your understanding of certain basic things:
A straight conductor of length ‘L’ carrying a current ‘I’ is bent in the form of a semicircle. The magnetic flux density (in tesla) at the centre of the semicircle is approximately
(a) 10-6I/L (b) 10-7I/L (c) 10-6π2I/L (d) 107π2I/L (e) π2I/L
The magnetic field at the centre of a single turn circular current carrying coil is μ0I/2r and hence the magnetic field at the centre of a semicircle is μ0I/4r. The radius of the semicircle obtained by bending the straight conductor of length L is, r = L/π. On substituting this value, the magnetic field at the centre of the semicircle is μ0πI/ 4L. Since μ0 = 4π ×10-7, the field is 10-7π2 I/L. But π2 is approximately equal to 10 so that the answer is 10-6I/L [Option (a)].
Let us now consider the following M.C.Q. which appeared in the Karnataka Common Entrance test paper of 2005:
The electrons in the beam of a television tube move horizontally from south to north. The vertical component of the earth’s magnetic field points down. The electron is deflected towards
(a) west (b) no deflection (c) east (d) north to south.
It will be convenient to use Fleming’s left hand rule here. You have to hold the fore-finger, middle finger and thumb of your left hand along mutually perpendicular directions so that the fore-finger points along the magnetic field and the middle finger along the direction of the conventional current (and hence opposite to the direction of the electron). The thumb will then give you the direction of deflection, which you can easily obtain as along the east [Option (c)]. You will find more multiple choice questions (with solution) at physicsplus: Magnetic Force on Moving Charges
An electron moves in a circular orbit with a uniform speed ‘v’. It produces a magnetic field ‘B’ at the centre of the circle. The radius of the circle is proportional to
(a) √(B/v) (b) B/v (c) √(v/B) (d) v/B
The electron, moving along the circular path is equivalent to a current I = e/T = ev/(2πr) where ‘e’ is the electronic charge, T is the period, ‘v’ is the speed and ‘r’ is the radius of the circular motion. The magnetic field at the centre of the circle, B = μ0I/2r = μ0ev/4πr2 so that r α √(v/B). The correct option is (c).
Now, consider the following simple question. (Be careful however, so that you won’t pick out a wrong answer):
A long straight conductor P carries a current ‘I’ while another parallel long straight conductor Q carries a current 2I in the same direction. The magnetic field midway between them is B. If the current 2I in the conductor Q is switched off, the magnetic field midway between the conductors will be
(a) B/3 (b) B (c) –B (d) –B/2 (e) 2B
The magnetic fields produced by the conductors are in opposition in the region between them. The field produced by Q is twice the field produced by P. If the field produced by Q is 2B, the field produced by P is –B. This is indeed the case, since the net field is B when both currents are switched on. When the current 2I in Q is switched off, the field due to Q alone is present. The answer therefore is –B [Option (c)].
The following question also is simple. It is designed to check your understanding of certain basic things:
A straight conductor of length ‘L’ carrying a current ‘I’ is bent in the form of a semicircle. The magnetic flux density (in tesla) at the centre of the semicircle is approximately
(a) 10-6I/L (b) 10-7I/L (c) 10-6π2I/L (d) 107π2I/L (e) π2I/L
The magnetic field at the centre of a single turn circular current carrying coil is μ0I/2r and hence the magnetic field at the centre of a semicircle is μ0I/4r. The radius of the semicircle obtained by bending the straight conductor of length L is, r = L/π. On substituting this value, the magnetic field at the centre of the semicircle is μ0πI/ 4L. Since μ0 = 4π ×10-7, the field is 10-7π2 I/L. But π2 is approximately equal to 10 so that the answer is 10-6I/L [Option (a)].
Let us now consider the following M.C.Q. which appeared in the Karnataka Common Entrance test paper of 2005:
The electrons in the beam of a television tube move horizontally from south to north. The vertical component of the earth’s magnetic field points down. The electron is deflected towards
(a) west (b) no deflection (c) east (d) north to south.
It will be convenient to use Fleming’s left hand rule here. You have to hold the fore-finger, middle finger and thumb of your left hand along mutually perpendicular directions so that the fore-finger points along the magnetic field and the middle finger along the direction of the conventional current (and hence opposite to the direction of the electron). The thumb will then give you the direction of deflection, which you can easily obtain as along the east [Option (c)]. You will find more multiple choice questions (with solution) at physicsplus: Magnetic Force on Moving Charges
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