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Chapter 13: Problem 1955

Unit of magnetic Flux density is (a) Tesla (b) [(Weber) \(/\) (meter) \(\left.^{2}\right]\) (c) [(Newton) \(/\) (Amp - meter)] (d) All of the above

Short Answer

Expert verified
The unit of magnetic flux density is (d) All of the above, as all given options are equivalent to Tesla (T).

Step by step solution

01

Option A: Tesla

Tesla (T) is the SI unit of magnetic flux density. It is a widely known unit to express the strength of a magnetic field and relates the magnetic field to the force acting on a charged particle. Therefore, Tesla could be a possible correct option.
02

Option B: [(Weber) \(/\) (meter) \(\left.^{2}\right]\)

Weber (Wb) is the unit for magnetic flux, and meter squared (m^2) is the unit for area. When we divide Weber by meter squared, it represents the magnetic flux distributed over an area, which is another way of describing the magnetic field strength. Coincidentally, this is equivalent to Tesla, meaning option (b) is correct as well.
03

Option C: [(Newton) \(/\) (Amp - meter)]

Newton (N) is the unit of force, while Ampere (A) is the unit for electric current and meter (m) for length. The expression \(\frac{\text{Newton}}{\text{Amp} \cdot \text{meter}}\) describes the amount of force experienced by a charged particle per unit current and length when it is placed in a magnetic field. This is the expression for magnetic field strength, which is again equivalent to Tesla. This indicates that option (c) is also correct.
04

Conclusion

As all the given options are equivalent to the unit Tesla (T), the correct answer for the unit of magnetic flux density is: (d) All of the above

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Most popular questions from this chapter

A Galvanometer of resistance \(15 \Omega\) is connected to a battery of 3 volt along with a resistance of \(2950 \Omega\) in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be (a) \(6050 \Omega\) (b) \(4450 \Omega\) (c) \(5050 \Omega\) (d) \(5550 \Omega\)

A domain in a ferromagnetic substance is in the form of a cube of side length \(1 \mu \mathrm{m}\). If it contains \(8 \times 10^{10}\) atoms and each atomic dipole has a dipole moment of \(9 \times 10^{-24} \mathrm{~A} \mathrm{~m}^{2}\) then magnetization of the domain is \(\mathrm{A} \mathrm{m}^{-1}\) a) \(7.2 \times 10^{5}\) (b) \(7.2 \times 10^{3}\) (c) \(7.2 \times 10^{5}\) (d) \(7.2 \times 10^{3}\)

0: Due to 10 Amp of current flowing in a circular coil of \(10 \mathrm{~cm}\) radius, the mag. field produced at its centre is \(\pi \times 10^{-3}\) Tesla. The number of turns in the coil will be (a) 5000 (b) 100 (c) 50 (d) 25

A small bar magnet of moment \(\mathrm{M}\) is placed in a uniform field of \(\mathrm{H}\). If magnet makes an angle of \(30^{\circ}\) with field, the torque acting on the magnet is (a) \(\mathrm{MH}\) (b) \((\mathrm{MH} / 2)\) (c) \((\mathrm{MH} / 3)\) (d) \((\mathrm{MH} / 4)\)

Two wires of same length are shaped into a square and a circle. If they carry same current, ratio of the magnetic moment is (a) \(2: \pi\) (b) \(\pi: 2\) (c) \(\pi: 4\) (d) \(4: \pi\)
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