EXERCISE SHORT QUESTIONS
CHAPTER # 14: ELECTROMAGNETISM
Q # 1. A plane conducting loop is located in a uniform magnetic field that is directed along the x-axis. For what orientations of the loop, is the flux maximum? For what orientation, is the flux minimum?}
Ans. The magnetic flux through a conducting loop can be find out by the expression:
\[\mathrm{\Delta}\varphi = \mathbf{B.A =}BA\cos\theta\]
Here \textbf{B} is the magnetic field strength and A is vector area whose direction is perpendicular to the plane of the loop.
Case 1. When vector area of the conducting loop is in the direction of magnetic field strength i.e., \(\theta = 0{^\circ}\), then the magnetic flux:
\[\mathrm{\Delta}\varphi = \mathbf{B.A =}BA\cos{0{^\circ}} = BA\)as\(\cos{0{^\circ}} = 1\]
Thus the magnetic flux through the coil is maximum, when the vector area of the conducting loop is parallel to magnetic field strength.
Case 2. When vector area of the conducting loop is perpendicular to magnetic field strength i.e., \(\theta = 90{^\circ}\), then the magnetic flux:
\(\mathrm{\Delta}\varphi = \mathbf{B.A =}BA\cos{90{^\circ}} = BA\)as\(\cos{90{^\circ}} = 0\)
Thus the magnetic flux through the coil is minimum, when the vector area of the conducting loop is perpendicular to magnetic field strength.
Q # 2. A current in a conductor produce a magnetic field, which can be calculated using Ampere's Law. Since current is defined as the rate of flow of charge. What can you conclude about the magnetic field due to stationary charges? What about moving charges?
Ans. A stationary charges cannot produce any magnetic field. In case of stationary charges, the rate of flow of charges is zero( i.e. current = 0), so there will be no magnetic field.
As the moving charges produce current, so the magnetic field produced around the path of its motion similar to the magnetic field produced around a current carrying conductor.
Q # 3. Describe the charge in the magnetic field inside a solenoid carrying steady current I, if (a) the length of the solenoid is doubled but the number of turns remains the same and (b) the number of turns are doubled, but the length remains the same.
Ans. The magnetic field strength \textbf{B} inside a current carrying conductor can be find out by the expression:
\[B\mathbf{=}\frac{\mu_{0}NI}{\mathbf{L}}\]
where I is the current flowing through conductor and N is the number of turn of solenoid having length L. Thus
When Length of solenoid is doubled by keeping the number of turns constant, then magnetic field
strength:
\[B'\mathbf{=}\frac{\mu_{0}NI}{2L}\mathbf{\Longrightarrow}B'\mathbf{=}\frac{B}{2}\]
Thus on doubling the length of solenoid by keeping the turns constant, the magnetic field strength becomes one half of its original value.
When number of turns of solenoid is doubled by keeping the length of solenoid constant, then magnetic field strength:
\[B''\mathbf{=}\frac{\mu_{0}(2N)I}{L}\mathbf{\Longrightarrow}B''\mathbf{=}2B\]
Thus on doubling the number of turns of solenoid by keeping its length constant, the magnetic field strength becomes doubled of its original value.
Q # 4. At a given instant, a proton moves in the positive x-direction in the region where there is magnetic field in the negative z-direction. What is the direction of the magnetic force? Will the proton continue to move in the positive x-direction? Explain.}
Ans. As the proton is moving in the positive x-direction and magnetic field is directed into the plane of paper, then the magnetic force on proton can be find out using expression:
\[\mathbf{F} = q\left( \mathbf{v \times B} \right)\]
According to right hand rule, the magnetic force is directed along y-axis.
No, the proton will not continue to move in the positive x-direction. Since the magnetic force is acting at the right angle to motion of conductor, therefore it will move along a circular path in xy-plane.
Q # 5. Two charged particles are projected into a region where there is a magnetic field perpendicular to their velocities. If the charges are deflected in opposite directions, what can you say about them?
Ans. When a charge particle is projected in a magnetic field, it will experience the magnetic force given by:
\[\mathbf{F} = q\left( \mathbf{v \times B} \right)\]
The magnetic force is a deflecting force. Thus if the charged particles are deflected in opposite direction, then particles are oppositely charged. i.e., one particle is positively charged and the other is negatively charged.
Q # 6. Suppose that a charge q is moving in a uniform magnetic field with a velocity v. Why is there no work done by the magnetic force that acts on the charge?
Ans. The magnetic force on a charge particle will act normal to the direction of motion of the particle, so the work done by the force is given by:
\[W\mathbf{= F.d =}Fd\cos\theta\]
Where \(\theta\) is the angle between the magnetic force and displacement of charge particle. For present case: \(\theta = 90{^\circ}\). Therefore:
\[W\mathbf{=}Fd\cos{90{^\circ}} = 0\]
Thus we can say that magnetic force is a deflecting force and it cannot do any work.
Q # 7. If a charge particle moves in a straight line through some region of space, can you say that the magnetic field in the region is zero.
Ans. The magnitude of magnetic force on a charge particle can be expressed as:
\[F = qvB\sin\theta\]
Where \(\theta\ \)is the angle between \textbf{B} and \textbf{v}. So if the particle moves in a straight line through some region of space then it means that the charge particle is not experiencing magnetic force which might be due to one of the following reasons:
Magnetic field strength B in the region is zero
Magnetic field is parallel or anti-parallel to the direction of motion.
Q # 8. Why does the picture on a TV screen become distorted when a magnet is brought near the screen?
Ans. The picture on a TV is formed when moving electrons strike the florescent screen. As magnet is brought close to the TV screen, the path of electrons is distorted due to the magnetic force on them. So the picture on the screen of TV is distorted.
Q # 9. Is it possible to orient a current loop in a uniform magnetic field such that the loop will not tend to rotate? Explain.
Ans. A current carrying loop when placed in magnetic field will experience a torque given by:
\[\tau = BINA\cos\alpha\]
Where B is the magnetic field strength, I is current flowing through coil, N is number of turns in a coil, A is the area of the coil and \(\alpha\) is the angle between plane of the coil and magnetic field.
It is clear from expression that when plane of the coil makes and angle of \(90{^\circ}\) with magnetic field, the torque on the coil will be zero. In this condition, the coil will not tend to rotate.
Q # 10. How can a current loop be used to determine the presence of a magnetic field in a given region of space?
Ans. When a current carrying loop is placed in a uniform magnetic field, a torque is produced in the loop is given by:
\[\tau = BINA\cos\alpha\]
If the loop is deflected in a given region, then it confirms the presence of magnetic field, otherwise not.
Q # 11. How can you use a magnetic field to separate isotopes of chemical element?}
Ans. If the ions of isotopes of an element are projected in a magnetic field of known strength B, the ions move in circular path of radius r. The e/m of the ion is given by the expression:
\[\frac{e}{m} = \frac{v}{Br}\mathbf{\Longrightarrow}r = \frac{v}{B} \times \frac{m}{e}\]
If v, B and e of the ions are constant, then
\[r \propto m\]
So the ions of different mass will have different radii of curvature and hence they can be separated in magnetic field.
Q # 12. What should be the orientation of a current carrying coil in a magnetic field so that torque acting upon the coil is (a) maximum (b) minimum?
Ans. A current carrying loop when placed in magnetic field will experience a torque given by:
\[\tau = BINA\cos\alpha\]
Where B is the magnetic field strength, I is current flowing through coil, N is number of turns in a coil, A is the area of the coil and \(\alpha\) is the angle between plane of the coil and magnetic field.
When plane of the coil is parallel to magnetic field, \(\alpha = 0\) and the torque acting on the coil will be maximum given by:
\(\tau = BINA\cos{0{^\circ}} = BINA\)
When plane of the coil is perpendicular to magnetic field, \(\alpha = 90{^\circ}\) and the torque acting on the coil will be minimum, given by:
\(\tau = BINA\cos{90{^\circ}} = 0\).
Q # 13. A loop of wire is suspended between the pools of a magnet with its plane parallel to the pole faces. What happens if a direct current is put through the coil? What happens if an alternating current is used instead?
Ans. As the plane of the coil is parallel to the pole faces, therefore, it is perpendicular to the magnetic field, i.e. α = 90°. Torque acting on coil
\(\mathbf{\tau = BINA}\mathbf{\cos}\mathbf{90{^\circ}}\mathbf{= 0}\).
Therefore, for both A.C. and D.C., the coil will not Rotate
Q # 14. Why the resistance of an ammeter should be very low?}
Ans. An ammeter is connected in series with a circuit to measure the current. It is connected in series so that total current passing through the circuit should pass through it. If the resistance of the ammeter will be large, it will alter the current of the circuit to great extent and the measurement of current will not be accurate.
Q # 15. Why the voltmeter should have a very high resistance?}
Ans. A voltmeter is connected in parallel to the resistor to measure potential difference across it. It should have very high resistance so that practically, a very little current should pass through it and the current of the circuit should almost remain constant, so that it might measure the potential difference across a resistor accurately.
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