Every Book Back multiple-choice question from Magnetism and magnetic effects of electric current (12th Standard Physics, Samacheer Kalvi) — each with the correct option highlighted and a clear, worked explanation. Free to read in English and Tamil.
Q1
The magnetic field at the centre O of the following current loop is
- A. \(\frac{\mu_0 I}{4r} \otimes\)Correct
- B. \(\frac{\mu_0 I}{4r} \odot\)
- C. \(\frac{\mu_0 I}{2r} \otimes\)
- D. \(\frac{\mu_0 I}{2r} \odot\)
Explanation. The magnetic field at the center of a current-carrying loop is determined using the Biot-Savart law for circular arcs.
Q2
An electron moves in a straight line inside a charged parallel plate capacitor of uniform charge density \(\sigma\). The time taken by the electron to cross the parallel plate capacitor undeflected when the plates of the capacitor are kept under constant magnetic field of induction \(\vec{B}\) is
- A. \(\frac{\epsilon_0 \sigma}{elB}\)
- B. \(\frac{\epsilon_0 \sigma}{lB}\)
- C. \(\frac{\epsilon_0 \sigma}{lBe}\)
- D. \(\frac{\epsilon_0 l B}{\sigma}\)Correct
Explanation. The condition for undeflected motion requires the electric force to equal the magnetic force, allowing the calculation of velocity and subsequently the transit time.
Q3
A particle having mass m and charge q accelerated through a potential difference V. Find the force experienced when it is kept under perpendicular magnetic field \(\vec{B}\).
- A. \(2\sqrt{\frac{q^3 BV}{m}}\)
- B. \(qB\sqrt{\frac{2V}{m}}\)
- C. \(\sqrt{\frac{2q^3 B^2 V}{m}}\)Correct
- D. \(\sqrt{\frac{2q^3 BV^2}{m^3}}\)
Explanation. The velocity acquired by the charge from the accelerating potential is determined using kinetic energy conservation, and this velocity is used in the Lorentz force expression for a perpendicular field.
Q4
A circular coil of radius 5 cm and 50 turns carries a current of 3 ampere. The magnetic dipole moment of the coil is nearly
- A. \(1.0\text{ Am}^2\)
- B. \(1.2\text{ Am}^2\)Correct
- C. \(0.5\text{ Am}^2\)
- D. \(0.8\text{ Am}^2\)
Explanation. The magnetic dipole moment of a coil is the product of the number of turns, the current flowing through it, and the area of the circular loop.
Q5
A thin insulated wire forms a plane spiral of \(N = 100\) tight turns carrying a current \(I = 8\text{ mA}\) (milli ampere). The radii of inside and outside turns are \(a = 50\text{ mm}\) and \(b = 100\text{ mm}\) respectively. The magnetic induction at the centre of the spiral is
- A. \(5\text{ \mu T}\)
- B. \(7\text{ \mu T}\)Correct
- C. \(8\text{ \mu T}\)
- D. \(10\text{ \mu T}\)
Explanation. The magnetic induction at the center is calculated by integrating the field contributions from infinitesimal circular turns over the range of the spiral's inner and outer radii.
Q6
Three wires of equal lengths are bent in the form of loops. One of the loops is circle, another is a semi-circle and the third one is a square. They are placed in a uniform magnetic field and same electric current is passed through them. Which of the following loop configuration will experience greater torque ?
- A. CircleCorrect
- B. Semi-circle
- C. Square
- D. All of them
Explanation. Torque is proportional to the area enclosed by the loop, and for a fixed perimeter length, a circular shape encloses the largest possible area.
Q7
Two identical coils, each with N turns and radius R are placed coaxially at a distance R as shown in the figure. If I is the current passing through the loops in the same direction, then the magnetic field at a point P at a distance of R/2 from the centre of each coil is
- A. \(\frac{8\mu_0 NI}{5R}\)
- B. \(\frac{8\mu_0 NI}{5\sqrt{5}R}\)Correct
- C. \(\frac{8\mu_0 NI}{5R}\)
- D. \(\frac{4\mu_0 NI}{5R}\)
Explanation. The magnetic field is calculated by applying the formula for the magnetic field on the axis of a circular current loop and summing the contributions from both coils.
Q8
A wire of length \(l\) carrying a current \(I\) along the Y direction is kept in a magnetic field given by \(\vec{B} = \beta(\hat{i} + \hat{j} + \hat{k})\text{ T}\). The magnitude of Lorentz force acting on the wire is
- A. \(\sqrt{2}\beta Il\)Correct
- B. \(\frac{1}{\sqrt{2}}\beta Il\)
- C. \(2\beta Il\)
- D. \(\frac{1}{2}\beta Il\)
Explanation. The force on a current-carrying wire is found by taking the vector cross product of the current-length element with the magnetic field vector and calculating its magnitude.
Q9
A bar magnet of length \(l\) and magnetic moment \(p_m\) is bent in the form of an arc as shown in figure. The new magnetic dipole moment will be
- A. \(p_m\)
- B. \(\frac{3}{\pi}p_m\)Correct
- C. \(\frac{2}{\pi}p_m\)
- D. \(\frac{1}{2}p_m\)
Explanation. The new dipole moment is calculated using the updated straight-line distance between the poles, which is determined from the geometry of the circular arc.
Q10
A non-conducting charged ring carrying a charge of q, mass m and radius r is rotated about its axis with constant angular speed \(\omega\). Find the ratio of its magnetic moment with angular momentum is
- A. \(q/m\)
- B. \(2q/m\)
- C. \(q/2m\)Correct
- D. \(q/4m\)
Explanation. The ratio is determined by dividing the expression for the magnetic dipole moment of the rotating ring by the expression for its rotational angular momentum.
Q11
The BH curve for a ferromagnetic material is shown in the figure. The material is placed inside a long solenoid which contains 1000 turns/cm. The current that should be passed in the solenoid to demagnetize the ferromagnet completely is
- A. \(1.00\text{ mA}\)
- B. \(1.25\text{ mA}\)
- C. \(1.50\text{ mA}\)Correct
- D. \(1.75\text{ mA}\)
Explanation. The coercivity value is identified from the H-axis of the B-H loop, and the required current is calculated using the formula for the solenoid's magnetizing field.
Q12
Two short bar magnets have magnetic moments \(1.20\text{ Am}^2\) and \(1.00\text{ Am}^2\) respectively. They are kept on a horizontal table parallel to each other with their north poles pointing towards south. They have a common magnetic equator and are separated by a distance of 20.0 cm. The value of the resultant horizontal magnetic induction at the mid-point O of the line joining their centres is (Horizontal components of Earth’s magnetic induction is \(3.6 \times 10^{-5}\text{ Wb m}^{-2}\))
- A. \(3.60 \times 10^{-5}\text{ Wb m}^{-2}\)
- B. \(3.5 \times 10^{-5}\text{ Wb m}^{-2}\)
- C. \(2.56 \times 10^{-4}\text{ Wb m}^{-2}\)Correct
- D. \(2.2 \times 10^{-4}\text{ Wb m}^{-2}\)
Explanation. The resultant field is the algebraic sum of the horizontal component of the Earth's magnetic field and the equatorial magnetic fields produced by each bar magnet.
Q13
The vertical component of Earth’s magnetic field at a place is equal to the horizontal component. What is the value of angle of dip at this place?
- A. \(30^\circ\)
- B. \(45^\circ\)Correct
- C. \(60^\circ\)
- D. \(90^\circ\)
Explanation. The angle of dip is the angle whose tangent is given by the ratio of the vertical component to the horizontal component of the magnetic field.
Q14
A flat dielectric disc of radius R carries an excess charge on its surface. The surface charge density is \(\sigma\). The disc rotates about an axis perpendicular to its plane passing through the centre with angular velocity \(\omega\). Find the magnitude of the torque on the disc if it is placed in a uniform magnetic field whose strength is B which is directed perpendicular to the axis of rotation
- A. \(\frac{1}{4}\sigma\omega\pi B R^4\)
- B. \(\frac{1}{2}\sigma\omega\pi B R^4\)
- C. \(\frac{1}{4}\sigma\omega\pi B R^3\)
- D. \(\frac{1}{4}\sigma\omega\pi B R^4\)Correct
Explanation. The torque is calculated by determining the magnetic moment of the rotating charged disc and then applying the formula for torque in a uniform magnetic field.
Q15
The potential energy of magnetic dipole whose dipole moment is \(\vec{p}_m = (-0.5\hat{i} + 0.4\hat{j})\text{ Am}^2\) kept in uniform magnetic field \(\vec{B} = 0.2\hat{i}\text{ T}\)
- A. \(-0.1\text{ J}\)
- B. \(-0.8\text{ J}\)
- C. \(0.1\text{ J}\)Correct
- D. \(0.8\text{ J}\)
Explanation. The potential energy is the negative dot product of the dipole moment vector and the uniform magnetic field vector.
About these Magnetism and magnetic effects of electric current questions
These are the Book Back multiple-choice questions for Magnetism and magnetic effects of electric current from the Tamil Nadu State Board (Samacheer Kalvi) 12th Standard Physics syllabus. Each question shows the correct option and an original, step-by-step explanation so you understand the method, not just the answer. Use the answer key above to jump to any question, then take the practice test to check yourself under exam-like conditions.
Frequently asked questions
How many MCQs are there in Magnetism and magnetic effects of electric current?
This chapter has 15 book-back multiple-choice questions, each with the correct answer and a step-by-step explanation.
Are these 12th Standard Physics MCQs free to practise online?
Yes. Every question, answer and explanation here is free, and you can also take them as a timed practice test.
Where can I find the Magnetism and magnetic effects of electric current book-back answers?
The correct option for each question is highlighted on this page with a worked explanation, plus a quick answer-key summary at the top.
