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Chapter 13: Problem 1916

A conducting rod of length \(\ell\) [cross-section is shown] and mass \(\mathrm{m}\) is moving down on a smooth inclined plane of inclination \(\theta\) with constant speed v. A vertically upward mag. field \(\mathrm{B}^{-}\) exists in upward direction. The magnitude of mag. field \(B^{-}\) is(a) $[(\mathrm{mg} \sin \theta) /(\mathrm{I} \ell)]$ (b) \([(\mathrm{mg} \cos \theta) /(\mathrm{I} \ell)]\) (c) \([(\mathrm{mg} \tan \theta) /(\mathrm{I} \ell)]\) (d) \([(\mathrm{mg}) /(\mathrm{I} \ell \sin \theta)]\)

Short Answer

Expert verified
The magnitude of the magnetic field B is given by: \( B = \frac{mg\sin\theta}{I\ell} \)

Step by step solution

01

Calculate gravitational force acting on the rod

The gravitational force acting on the rod can be resolved into two components: one perpendicular to the inclined plane (mgcosθ) and one parallel to it (mgsinθ). The force parallel to the inclined plane (mgsinθ) causes the rod to slide down.
02

Calculate the magnetic force acting on the rod

According to Faraday's law of electromagnetic induction, the conducting rod experiences an induced electromotive force (EMF) as it moves in the magnetic field B. This induces an electric current, I, through the rod. Due to the magnetic field, this electric current exerts a magnetic force on the rod. The magnetic force, F, acting on the rod can be given by the equation: \( F = I\ell B \) where F is the magnetic force, I is the induced current, 𝑙 is the length of the rod, and B is the magnetic field.
03

Apply equilibrium conditions for the rod moving with constant speed

Since the rod is moving at a constant speed, the net force acting on the rod along the direction of motion is zero. Therefore, we have: Governing_force (Gravity) = Governing_force (Magnetic) or \( mgsin\theta = I \ell B \)
04

Solve for the magnetic field, B

Now, we will rearrange the equation derived in step 3 to find B: \( B = \frac{mgsin\theta}{I\ell} \) Comparing the given options with our derived expression for B, we can see that the correct answer is given by option (a): \( B = \frac{mg\sin\theta}{I\ell} \)

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Most popular questions from this chapter

A conducting circular loop of radius a carries a constant current I. It is placed in a uniform magnetic field \(\mathrm{B}^{-}\), such that \(\mathrm{B}^{-}\) is perpendicular to the plane of the Loop. The magnetic force acting on the Loop is (a) \(\mathrm{B}^{-} \operatorname{Ir}\) (b) \(\mathrm{B}^{-} \mathrm{I} \pi \mathrm{r}^{2}\) (c) Zero (d) BI \((2 \pi \mathrm{r})\)

The effective length of a magnet is \(31.4 \mathrm{~cm}\) and its pole strength is \(0.5 \mathrm{~A} \mathrm{~m}\). The magnetic moment, if it is bent in the form of a semicircle will be Amp.m \(^{2}\) (a) \(0.1\) (b) \(0.01\) (c) \(0.2\) (d) \(1.2\)

A thin magnetic needle oscillates in a horizontal plane with a period \(\mathrm{T}\). It is broken into n equal parts. The time period of each part will be (a) \(\mathrm{T}\) (b) \(\mathrm{n}^{2} \mathrm{~T}\) (c) \((\mathrm{T} / \mathrm{n})\) (d) \(\left(\mathrm{T} / \mathrm{n}^{2}\right)\)

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|r} Column-I & Column - II \end{tabular} (A) Magnetic field induction due to Current 1 through straight conductor at a perpendicular distance \(\mathrm{r}\). (B) Magnetic field induction at (Q) \(\left[\left(\mu_{0} \mathrm{I}\right) /(4 \pi \mathrm{r})\right]\) the centre of current \((1)\) carrying Loop of radius (r) (C) Magnetic field induction at the (R) \(\left[\mu_{0} /(4 \pi)\right](2 \mathrm{I} / \mathrm{r})\) axis of current (1) carrying coil of radius (r) at a distance (r) from centre of coil (D) Magnetic field induction at the (S) \(\left[\mu_{0} /(4 \sqrt{2})\right](\mathrm{L} / \mathrm{r})\) at the centre due to circular arc of length \(\mathrm{r}\) and radius (r) carrying current (I) (a) $\mathrm{A} \rightarrow \mathrm{R} ; \mathrm{B} \rightarrow \mathrm{S} ; \mathrm{C} \rightarrow \mathrm{P} ; \mathrm{D} \rightarrow \mathrm{Q}$ (b) $\mathrm{A} \rightarrow \mathrm{R} ; \mathrm{B} \rightarrow \mathrm{P} ; \mathrm{C} \rightarrow \mathrm{S} ; \mathrm{D} \rightarrow \mathrm{Q}$ (c) $\mathrm{A} \rightarrow \mathrm{P} ; \mathrm{B} \rightarrow \mathrm{Q} ; \mathrm{C} \rightarrow \mathrm{S} ; \mathrm{D} \rightarrow \mathrm{R}$ (d) $\mathrm{A} \rightarrow \mathrm{Q} ; \mathrm{B} \rightarrow \mathrm{P} ; \mathrm{C} \rightarrow \mathrm{R} ; \mathrm{D} \rightarrow \mathrm{S}$

The angles of dip at two places are \(30^{\circ}\) and \(45^{\circ}\). The ratio of horizontal components of earth's magnetic field at the two places will be (a) \(\sqrt{3}: \sqrt{2}\) (b) \(1: \sqrt{2}\) (c) \(1: 2\) (d) \(1: \sqrt{3}\)
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