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Chapter 13: Problem 1989

The coercively of a bar magnet is \(100 \mathrm{~A} / \mathrm{m}\). It is to be diamagnetism by placing it inside a solenoid of length \(100 \mathrm{~cm}\) and number of turns 50 . The current flowing through the solenoid will be (a) \(4 \mathrm{~A}\) (b) \(2 \mathrm{~A}\) (c) \(1 \mathrm{~A}\) (d) Zero

Short Answer

Expert verified
The current flowing through the solenoid for diamagnetism is \(2 \mathrm{~A}\), which corresponds to option (b).

Step by step solution

01

Write down the given values

The given values are: Coercivity of the bar magnet \(H_{c} = 100 \mathrm{~A}\) \(/\) \(\mathrm{m}\), Length of the solenoid \(l = 100 \mathrm{~cm}\) \(= 1 \mathrm{~meter}\) (converting to meters), Number of turns of the solenoid \(N = 50\).
02

Use the coercive field formula

The coercive field formula is \(H_{c} = NI / l\). We can solve this formula for the current \(I\): \(I = \frac{H_{c} \cdot l}{N}\)
03

Substitute the given values in the formula

Now insert the given values into the formula to find the current \(I\): \(I = \frac{(100 \mathrm{~A} / \mathrm{m}) \cdot (1 \mathrm{~m})}{50}\)
04

Calculate the current

Simplify the expression to get the current \(I\): \(I = \frac{100 \mathrm{~A}}{50} = 2 \mathrm{~A}\) The current flowing through the solenoid for diamagnetism is \(2 \mathrm{~A}\), which corresponds to option (b).

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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 particle of mass \(\mathrm{m}\) and charge q moves with a constant velocity v along the positive \(\mathrm{x}\) -direction. It enters a region containing a uniform magnetic field B directed along the negative \(z\) -direction, extending from \(x=a\) to \(x=b\). The minimum value of required so that the particle can just enter the region \(\mathrm{x}>\mathrm{b}\) is (a) \([(\mathrm{q} \mathrm{bB}) / \mathrm{m}]\) (b) \(q(b-a)(B / m)\) (c) \([(\mathrm{qaB}) / \mathrm{m}]\) (d) \(q(b+a)(B / 2 m)\)

The forces existing between two parallel current carrying conductors is \(\mathrm{F}\). If the current in each conductor is doubled, then the value of force will be (a) \(2 \mathrm{~F}\) (b) \(4 \mathrm{~F}\) (c) \(5 \mathrm{~F}\) (d) \((\mathrm{F} / 2)\)

A 2 Mev proton is moving perpendicular to a uniform magnetic field of \(2.5\) tesla. The force on the proton is (a) \(3 \times 10^{-10} \mathrm{~N}\) (b) \(70.8 \times 10^{-11} \mathrm{~N}\) (c) \(3 \times 10^{-11} \mathrm{~N}\) (d) \(7.68 \times 10^{-12} \mathrm{~N}\)

Magnetic intensity for an axial point due to a short bar magnet of magnetic moment \(\mathrm{M}\) is given by (a) \(\left(\mu_{0} / 4 \pi\right)\left(\mathrm{M} / \mathrm{d}^{3}\right)\) (b) \(\left(\mu_{0} / 4 \pi\right)\left(\mathrm{M} / \mathrm{d}^{2}\right)\) (c) \(\left(\mu_{0} / 2 \pi\right)\left(\mathrm{M} / \mathrm{d}^{3}\right)\) (d) \(\left(\mu_{0} / 2 \pi\right)\left(\mathrm{M} / \mathrm{d}^{2}\right)\)
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