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Chapter 15: Problem 8

Earth's magnetic induction at a certain point is \(7 \times 10^{-5}\) \(\mathrm{Wb} / \mathrm{m}^{2}\). This field is to be annulled by the magnetic induction at the centre of a circular conducing loop \(5.0 \mathrm{~cm}\) in radius. The required current is (A) \(0.056 \mathrm{~A}\) (B) \(6.5 \mathrm{~A}\) (C) \(5.6 \mathrm{~A}\) (D) \(12.8 \mathrm{~A}\)

Short Answer

Expert verified
The required current to annul Earth's magnetic field at the center of a circular conducting loop with a radius of 5.0 cm is approximately \(5.6 \mathrm{A}\).

Step by step solution

01

Analyze the given data

First, we have the Earth's magnetic induction (\(B_{earth}\)) which is equal to \(7 \times10^{-5} \mathrm{Wb/m^2}\). The radius of the circular conducting loop (\(r\)) is given as \(5.0 \mathrm{cm}\) or \(0.05\mathrm{m}\).
02

Find the magnetic field at the center of the loop

According to Ampere's law, for a circular loop, we can find the magnetic field at the center (\(B_{loop}\)) using the following formula: \(B_{loop} = \frac{\mu_0 I}{2r}\) Where: \(B_{loop}\) = Magnetic field at the center of the loop, \(\mu_0\) = permeability of free space \(\approx 4\pi \times10^{-7} \mathrm{Wb/(A m)}\), \(I\) = Current in the loop, \(r\) = radius of the loop.
03

Cancel Earth's magnetic field with the loop's magnetic field

To cancel Earth's magnetic field at the center of the loop, we must adjust the current in the loop such that \(B_{loop} = B_{earth}\). Therefore, we can write the equation as: \(\frac{\mu_0 I}{2r} = B_{earth}\)
04

Solve for the required current, I

Now, we need to solve for the current (I) required in the loop to cancel the earth's magnetic field. Rearranging the above equation for I, we get: \(I = \frac{2r \times B_{earth}}{\mu_0}\) Substituting the known values: \(I = \frac{2 \times 0.05\mathrm{m} \times 7 \times10^{-5} \mathrm{Wb/m^2}}{4\pi \times10^{-7} \mathrm{Wb/(A m)}}\)
05

Calculate the value of the current

Finally, calculating the current value, we get: \(I \approx 5.6 \mathrm{A}\) Thus, the correct answer is (C) 5.6 A.

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Most popular questions from this chapter

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) \(\frac{3 \mu_{0}}{2 \pi}\)

Two particles \(X\) and \(Y\) having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii \(R_{1}\) and \(R_{2}\), respectively. The ratio of masses of \(X\) and \(Y\) is (A) \(\left(\frac{R_{1}}{R_{2}}\right)^{1 / 2}\) (B) \(\frac{R_{2}}{R_{1}}\) (C) \(\left(\frac{R_{1}}{R_{2}}\right)^{2}\) (D) \(\left(\frac{R_{1}}{R_{2}}\right)\)

A long straight wire carrying a current of \(30 \mathrm{~A}\) is placed in an external uniform magnetic field of magnitude \(4 \times\) \(10^{-4} \mathrm{~T}\). The magnetic field is acting parallel to the direction of current. The magnitude of the resultant magnetic field in tesla at a point \(2.0 \mathrm{~cm}\) away from the wire is (A) \(10^{-4}\) (B) \(3 \times 10^{-4}\) (C) \(5 \times 10^{-4}\) (D) \(6 \times 10^{-4}\)

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.

A particle of charge \(q\) and mass \(m\) starts moving from origin under the action of an electric field \(\vec{E}=E_{0} \hat{i}\) and magnetic field \(\vec{B}=B_{0} \hat{k}\). Its velocity at \((x, 0,0)\) is \(6 \hat{i}+8 \hat{j}\). The value of \(x\) is (A) \(\frac{25 m}{q E_{0}}\) (B) \(\frac{100 m}{q B_{0}}\) (C) \(\frac{50 m}{q E_{0}}\) (D) \(\frac{14 m}{q E_{0}}\)
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