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Two concentric coils each of radius equal to \(2 \pi \mathrm{cm}\) are placed at right angles to each other. 3 Amp and \(4 \mathrm{Amp}\) are the currents flowing in each coil respectively. The magnetic field intensity at the centre of the coils will be Tesla. (a) \(5 \times 10^{-5}\) (b) \(7 \times 10^{-5}\) (c) \(12 \times 10^{-5}\) (d) \(10^{-5}\)

Short Answer

Expert verified
The magnetic field intensity at the center of the concentric coils will be \(B_{net} = \sqrt{B_1^2 + B_2^2}\), where \(B_1 = \frac{4\pi \times 10^{-7} (3)}{2(2\pi \times 10^{-2})}\) and \(B_2 = \frac{4\pi \times 10^{-7} (4)}{2(2\pi \times 10^{-2})}\). After calculating \(B_{net}\), compare the value with the given options and choose the correct one.

Step by step solution

01

Calculate the magnetic field intensity produced by each coil

We will use the Biot-Savart Law to find the magnetic field intensity produced by each coil. The formula for the magnetic field intensity (B) at the center of a circular coil is given by: \[B = \frac{\mu_0 I}{2R}\] where \(\mu_0 = 4\pi \times 10^{-7} Tm/A\) (permeability of free space), \(I\) is the current flowing in the coil, and \(R\) is the radius of the coil. Using the given information, For Coil 1: \(I_1 = 3 A\), \(R = 2\pi cm = 2\pi \times 10^{-2} m\) For Coil 2: \(I_2 = 4 A\), \(R = 2\pi cm = 2\pi \times 10^{-2} m\) Calculate the magnetic field intensity produced by each coil: For Coil 1: \(B_1 = \frac{4\pi \times 10^{-7} (3)}{2(2\pi \times 10^{-2})}\) For Coil 2: \(B_2 = \frac{4\pi \times 10^{-7} (4)}{2(2\pi \times 10^{-2})}\)
02

Find the vector sum of the magnetic field intensities

Since the coils are placed at right angles to each other, we can think of the net magnetic field intensity as the hypotenuse of a right triangle, where the sides are the individual magnetic field intensities produced by the coils. We can find the net magnetic field intensity (B_net) using the Pythagorean theorem: \[B_{net} = \sqrt{B_1^2 + B_2^2}\] Substitute the values calculated in Step 1 and find \(B_{net}\).
03

Identify the correct option

Once we have calculated the net magnetic field intensity at the center of the concentric coils, compare the value with the options given in the problem and choose the closest option.

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Most popular questions from this chapter

In each of the following questions, Match column-I and column-II and select the correct match out of the four given choices.\begin{tabular}{l|l} Column - I & Column - II \end{tabular} (A) Biot-savart's law (P) Direction of magnetic field induction (B) Right hand thumb rule (Q) Magnitude of magnetic field induction (C) Fleming's left hand rule (R) Direction of induced current (D) Fleming's right hand rule (S) Direction of force due to a mag. field (a) $\mathrm{A} \rightarrow \mathrm{Q} ; \mathrm{B} \rightarrow \mathrm{P} ; \mathrm{C} \rightarrow \mathrm{R} ; \mathrm{D} \rightarrow \mathrm{S}$ (b) $\mathrm{A} \rightarrow \mathrm{Q} ; \mathrm{B} \rightarrow \mathrm{P} ; \mathrm{C} \rightarrow \mathrm{S} ; \mathrm{D} \rightarrow \mathrm{R}$ (c) $\mathrm{A} \rightarrow \mathrm{P} ; \mathrm{B} \rightarrow \mathrm{Q} ; \mathrm{C} \rightarrow \mathrm{R} ; \mathrm{D} \rightarrow \mathrm{S}$ (d) $\mathrm{A} \rightarrow \mathrm{P}: \mathrm{B} \rightarrow \mathrm{Q}: \mathrm{C} \rightarrow \mathrm{S} ; \mathrm{D} \rightarrow \mathrm{R}$

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) \(4: \pi\)

If a magnet of pole strength \(\mathrm{m}\) is divided into four parts such that the length and width of each part is half that of initial one, then the pole strength of each part will be (a) \((\mathrm{m} / 4)\) (b) \((\mathrm{m} / 2)\) (c) \((\mathrm{m} / 8)\) (d) \(4 \mathrm{~m}\)

The mag. field (B) at the centre of a circular coil of radius "a", through which a current I flows is (a) \(\mathrm{B} \propto \mathrm{c} \mathrm{a}\) (b) \(\mathrm{B} \propto(1 / \mathrm{I})\) (c) \(\mathrm{B} \propto \mathrm{I}\) (d) \(B \propto I^{2}\)

A straight wire of mass \(200 \mathrm{gm}\) and length \(1.5\) meter carries a current of 2 Amp. It is suspended in mid-air by a uniform horizontal magnetic field B. [take \(\left.\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right]\). The \(\mathrm{B}\) is (a) \(\overline{(2 / 3) \text { tes } 1 a}\) (b) \((3 / 2)\) tesla (c) \((20 / 3)\) tesla (d) (3/20) tesla

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