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Chapter 13: Problem 1918

A magnetic field existing in a region is given by $\mathrm{B}^{-}=\mathrm{B}_{0}[1+(\mathrm{x} / \ell)] \mathrm{k} \wedge . \mathrm{A}\( square loop of side \)\ell$ and carrying current I is placed with edges (sides) parallel to \(\mathrm{X}-\mathrm{Y}\) axis. The magnitude of the net magnetic force experienced by the Loop is (a) \(2 \mathrm{~B}_{0} \overline{\mathrm{I} \ell}\) (b) \((1 / 2) B_{0} I \ell\) (c) \(\mathrm{B}_{\circ} \mathrm{I} \ell\) (d) BI\ell

Short Answer

Expert verified
The correct answer is not among the given options. The net magnetic force experienced by the square loop is \(3 I \ell B_0\).

Step by step solution

01

Calculate the magnetic field at each side of the square loop

We are given the magnetic field in the region as \(B^{-}=B_0[1+(\frac{x}{\ell})]\,k\). The square loop has sides of length \(\ell\) and is placed parallel to the XY plane, which means the loop has vertices at (0,0), (0, \(\ell\)), (\(\ell\), \(\ell\)), and (\(\ell\), 0). We'll calculate the magnetic field at each side of the loop: Side 1: At x = 0, the magnetic field is \(B_1 = B_0 k\) Side 2: At x = \(\ell\), the magnetic field is \(B_2 = B_0 (1 + 1) k = 2 B_0 k\) Side 3: At x = \(\ell\), the magnetic field is \(B_3 = 2 B_0 k\), as in Side 2. Side 4: At x = 0, the magnetic field is \(B_4 = B_0 k\), as in Side 1.
02

Determine the magnetic force on each side of the loop

The magnetic force on each side can be determined using the formula \(F = I * l * B * sin(\angle)\). Since the angle between the magnetic field and the loop is 90 degrees, sin(90) = 1. So the magnetic forces on the four sides of the loop can be found using the following calculations: Force on Side 1: \(F_1 = I \ell B_1 = I \ell B_0\) Force on Side 2: \(F_2 = I \ell B_2 = I \ell (2 B_0)\) Force on Side 3: \(F_3 = 0\) (since the current is parallel to the magnetic field, the force is 0) Force on Side 4: \(F_4 = 0\) (similarly, the current is parallel to the magnetic field, the force is 0)
03

Sum up the magnetic forces for all sides of the loop

To find the net magnetic force experienced by the loop, we sum up the magnetic forces on all sides: Net magnetic force: \(F = F_1 + F_2 + F_3 + F_4 = I \ell B_0 + I \ell (2 B_0) = 3 I \ell B_0\) Now let's compare this result to the given options: a) \(2 \mathrm{~B}_{0} \overline{\mathrm{I} \ell}\) b) \((1 / 2) B_{0} I \ell\) c) \(\mathrm{B}_{\circ} \mathrm{I} \ell\) d) BI\ell None of the options matches our result (3I\(\ell\)B\(_{0}\)); there's a mistake in the given options. Nevertheless, we've shown how to calculate the net magnetic force experienced by the square loop.

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Most popular questions from this chapter

A long straight wire carrying current of \(30 \mathrm{Amp}\) is placed in an external uniform mag. field of induction \(4 \times 10^{-4}\) tesla. The mag. field is acting parallel to the direction of current. The magnitude of the resultant magnetic induction in tesla at a point \(2 \mathrm{~cm}\) away from the wire is tesla. (a) \(10^{-4}\) (c) \(5 \times 10^{-4}\) (b) \(3 \times 10^{-4}\) (d) \(6 \times 10^{-4}\)

Two iclentical short bar magnets, each having magnetic moment \(\mathrm{M}\) are placed a distance of \(2 \mathrm{~d}\) apart with axes perpendicular to each other in a horizontal plane. The magnetic induction at a point midway between them is. (a) $\sqrt{2}\left(\mu_{0} / 4 \pi\right)\left(\mathrm{M} / \mathrm{d}^{3}\right)$ (b) $\sqrt{3}\left(\mu_{0} / 4 \pi\right)\left(\mathrm{M} / \mathrm{d}^{3}\right)$ (c) $\sqrt{4}\left(\mu_{0} / 4 \pi\right)\left(\mathrm{M} / \mathrm{d}^{3}\right)$ (d) $\sqrt{5}\left(\mu_{0} / 4 \pi\right)\left(\mathrm{M} / \mathrm{d}^{3}\right)$

Read the assertion and reason carefully to mark the correct option out of the options given below. (A) If both assertion and reason are true and the reason is the correct explanation of the assertion. (B) If both assertion and reason are true but reason is not the correct explanation of the assertion, (C) If assertion is true but reason is false. (D) If the assertion and reason both the false. (E) If assertion is false but reason is true. Assertion: Diamagnetic materials can exhibit magnetism. Reason: Diamagnetic materials have permanent magnetic dipole moment.

An electron having mass \(9 \times 10^{-31} \mathrm{~kg}\), charge $1.6 \times 10^{-19} \mathrm{C}\( and moving with a velocity of \)10^{6} \mathrm{~m} / \mathrm{s}$ enters a region where mag. field exists. If it describes a circle of radius \(0.10 \mathrm{~m}\), the intensity of magnetic field must be Tesla (a) \(1.8 \times 10^{-4}\) (b) \(5.6 \times \overline{10^{-5}}\) (c) \(14.4 \times 10^{-5}\) (d) \(1.3 \times 10^{-6}\)

0: Due to 10 Amp of current flowing in a circular coil of \(10 \mathrm{~cm}\) radius, the mag. field produced at its centre is \(\pi \times 10^{-3}\) Tesla. The number of turns in the coil will be (a) 5000 (b) 100 (c) 50 (d) 25
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