Question

a stiff metal rod kept over two knife edges of length 1m. rod carries current of 16A. and rolls over rails without slipping due to uniform magnetic field of 0.5T perpendicular pointing downwards. rod starts from rest and attains speed of K/root5 when leaves rails. find value of K

Answer 1

Answer on Question#79283 - Physics - Electromagnetism

A stiff metal rod kept over two knife edges of length $L = 1 \, \text{m}$. Rod carries current of $I = 16 \, \text{A}$ and rolls over rails without slipping due to uniform magnetic field of $B = 0.5 \, \text{T}$ perpendicular pointing downwards. Rod starts from rest and attains speed of $K / \sqrt{5}$ when leaves rails. Find value of $K$.

**Solution:**

Let's denote the length of the rod as $l$, then the work done magnetic field is given by

$$A_{m} = I B l L$$

The translational kinetic energy:

$$E_{t} = \frac{1}{2} m v^{2},$$

where $m$ – mass of the rod, $v$ – linear velocity of the rod.

The rotational kinetic energy:

$$E_{r} = \frac{1}{2} I \omega^{2},$$

where $I$ – moment of inertia of the rod around its axis, $\omega$ – angular velocity of the rod.

The moment of inertia of a cylinder of mass $m$ and radius $r$ is given by

$$I = \frac{1}{2} m r^{2}$$

Since the rod doesn't slip, its linear and angular velocities are related as follows

$$v = \omega r$$

Total kinetic energy:

$$E_{k} = E_{t} + E_{r} = \frac{1}{2} m v^{2} + \frac{1}{2} I \omega^{2} = \frac{1}{2} m v^{2} + \frac{1}{2} \frac{1}{2} m r^{2} \left(\frac{v}{r}\right)^{2} = \frac{3}{4} m v^{2}$$

According to the law of conservation of energy the kinetic energy gained by the rod is equal to the work done by the magnetic field:

$$E_{k} = A_{m}$$

Thus

$$\frac{3}{4} m v^{2} = I B l L$$ $$v = \sqrt{\frac{4 I B l L}{3 m}} = \sqrt{\frac{20 I B l L}{3 m}} / \sqrt{5}$$

Therefore

$$K = \sqrt {\frac {20IBIL}{3m}}$$

For numerical answer the length of the rod $l$ and its mass are needed.

Answer: $K = \sqrt{\frac{20IBIL}{3m}}$.

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