Question

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

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

Step-by-step solution

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.

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.

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 \( mgsin\theta = I \ell B \)

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): \( B = \frac{mg\sin\theta}{I\ell} \)