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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

Two parallel long wires \(\mathrm{A}\) and B carry currents \(\mathrm{I}_{1}\) and \(\mathrm{I}_{2}\). \(\left(\mathrm{I}_{2}<\mathrm{I}_{1}\right)\) when \(\mathrm{I}_{1}\) and \(\mathrm{I}_{2}\) are in the same direction the mag. field at a point mid way between the wires is \(10 \mu \mathrm{T}\). If \(\mathrm{I}_{2}\) is reversed, the field becomes \(30 \mu \mathrm{T}\). The ratio \(\left(\mathrm{I}_{1} / \mathrm{I}_{2}\right)\) is (a) 1 (b) 2 (c) 3 (d) 4

In the case of bar magnet, lines of magnetic induction (a) Start from the North pole and end at the South pole (b) Run continuously through the bar and outside (c) Emerge in circular paths from the middle of the bar (d) Are produced only at the North pole like rays of light from a bulb.

The mag. field due to a current carrying circular Loop of radius $3 \mathrm{~cm}\( at a point on the axis at a distance of \)4 \mathrm{~cm}$ from the centre is \(54 \mu \mathrm{T}\) what will be its value at the centre of the LOOP. (a) \(250 \mu \mathrm{T}\) (b) \(150 \mu \mathrm{T}\) (c) \(125 \mu \mathrm{T}\) (d) \(75 \mu \mathrm{T}\)

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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) \(7.68 \times 10^{-12} \mathrm{~N}\)
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