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Chapter 15: Problem 58

A solenoid of \(1.5\) metre length and \(4.0 \mathrm{~cm}\) diameter posses 10 turn per \(\mathrm{cm}\). A current of 5 ampere is flowing through it. The magnetic induction at axis inside the solenoid is (A) \(2 \pi \times 10^{-3} \mathrm{~T}\) (B) \(2 \pi \times 10^{-5} \mathrm{~T}\) (C) \(4 \pi \times 10^{-2}\) gauss (D) \(2 \pi \times 10^{-5}\) gauss

Short Answer

Expert verified
The short answer is: (A) \(2 \pi \times 10^{-3} \mathrm{~T}\).

Step by step solution

01

Find the number of turns per meter

To find the number of turns per meter, multiply the given number of turns per cm by 100 (since there are 100 cm in a meter). The number of turns per cm is 10, so the number of turns per meter is \(10 \times 100 = 1000\).
02

Calculate the magnetic field inside the solenoid

Now, we can use the formula to find the magnetic field inside the solenoid: \(B = \mu_0 n I\). The permeability of free space, \(\mu_0\), is \(4\pi\times10^{-7} Tm/A\). We already know the number of turns per meter, \(n = 1000\), and the current, \(I = 5 A\). So, the magnetic field inside the solenoid is: \[ B = (4\pi\times10^{-7})(1000)(5) = 2\pi\times10^{-3} T. \]
03

Compare the calculated value to the given options

We found that the magnetic field inside the solenoid is \(2\pi\times10^{-3} T\), which matches option (A). Therefore, the correct answer is: (A) \(2 \pi \times 10^{-3} \mathrm{~T}\).

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Most popular questions from this chapter

Two identical conducting wires \(A O B\) and \(C O D\) are placed at right angles to each other. The wire \(A O B\) carries an electric current \(I_{1}\) and \(C O D\) carries a current \(I_{2} .\) The magnetic field on a point lying at a distance \(d\) from \(O\), in a direction perpendicular to the plane of the wires \(A O B\) and \(C O D\), will be given by [2007] (A) \(\frac{\mu_{0}}{2 \pi d}\left(I_{1}^{2}+I_{2}^{2}\right)\) (B) \(\frac{\mu_{0}}{2 \mu}\left(\frac{I_{1}+I_{2}}{d}\right)^{\frac{1}{2}}\) (C) \(\frac{\mu_{0}}{2 \pi d}\left(I_{1}^{2}+I_{2}^{2}\right)^{\frac{1}{2}}\) (D) \(\frac{\mu_{0}}{2 \pi d}\left(I_{1}+I_{2}\right)\)

An electron accelerated by a potential difference \(V=3.6 \mathrm{~V}\) first enters into a uniform electric field of a parallel-plate capacitor whose plates extend over a length \(l=6 \mathrm{~cm}\) in the direction of initial velocity. The electric field is normal to the direction of initial velocity and its strength varies with time as \(E=a \times t\), where \(a=\) \(3200 \mathrm{Vm}^{-1} \mathrm{~s}^{-1}\). Then the electron enters into a uniform magnetic field of induction \(B=\pi \times 10^{-9} \mathrm{~T}\). Direction of magnetic field is same as that of the electric field. Calculate pitch (in \(\mathrm{mm}\) ) of helical path traced by the electron in the magnetic field (Mass of electron, \(m=9 \times\) \(10^{-31} \mathrm{~kg}\) ). [Neglect the effect of induced magnetic field.]

Two long parallel wires \(P\) and \(Q\) are held perpendicular to the plane of the paper at a separation of \(5 \mathrm{~m}\). If \(P\) and \(Q\) carry currents of \(2.5 \mathrm{~A}\) and \(5 \mathrm{~A}\), respectively, in the same direction, then the magnetic field at a point midway between \(P\) and \(Q\) is (A) \(\frac{\mu_{0}}{\pi}\) (B) \(\frac{\sqrt{3} \mu_{0}}{\pi}\) (C) \(\frac{\mu_{0}}{2 \pi}\) (d) \(\frac{3 \mu_{0}}{2 \pi}\)

A charged particle enters a region which offers some resistance against its motion, and a uniform magnetic field exists in the region. The particle traces a spiral path as shown. Then (A) angular velocity of particle remains constant. (B) speed of particle decreases continuously. (C) total mechanical energy of the particle remains conserved. (D) net force on the particle is always perpendicular to its direction of motion.

When a long wire carrying a steady current is bent into a circular coil of one turn, the magnetic induction at its centre is \(B\). When the same wire carrying the same current is bent to form a circular coil of \(n\) turns of a smaller radius, the magnetic induction at the centre will be (A) \(B / n\) (B) \(n B\) (C) \(B / n^{2}\) (D) \(n^{2} B\)
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