Question

An athlete with a 3m long iron rod in hand runs towards the east with a speed of 30 kmph. The horizontal component of earth's magnetic field is $4\times {{10}^{-5}}Wb{{m}^{-2}}$ . If he runs with the rod in horizontal and vertical positions then the induced emf generated in the rod in two cases will be
A. Zero in vertical position and $1\times {{10}^{-3}}$ volt in horizontal position.
B. $1\times {{10}^{-3}}$ Volt in vertical position and zero volt in horizontal position
C. Zero in both positions
D. $1\times {{10}^{-3}}$Volt in both positions

Answer 1

Hint: Here we will use the Faraday’s law of electromagnetic induction and the equation of induced emf. In this question, we are asked to determine the induced emf in the rod for a particular case of an athlete when he runs with the rod in horizontal and vertical positions. So first we will calculate the induced emf of the rod and we know that in horizontal direction e=0.

Formula used:
Equation of induced emf:
$e=B\times v\times l$
Where,
B-Earth’s magnetic field
v-the speed
l-length of the rod

Complete answer:
This is one of the basic laws in electromagnetism. It states that the magnitude of induced emf is directly proportional to the rate of change of magnetic flux that cuts across the circuit. In both cases velocity is towards east with a magnitude of 30 kmph. We know that earth’s magnetic field is south to north.

In this question, if we consider in the first case an athlete is running towards east with a rod holding a rod in vertical direction, then the rod cuts the magnetic field of earth perpendicularly. Therefore, the rod experiences maximum induced emf.

Since magnetic field and velocity of the rod are perpendicular to each other, we can write it as:
velocity is given as kmph. To convert it to m/s ,multiply it simply with $\frac{5}{18}$
That is 1 km=1000m and $1 hr=60\times 60$
Therefore,
velocity of rod,
$v=30kmph=30\times \frac{1000}{60\times 60}=30\times \frac{5}{18}=\frac{25}{3}ms^{-1}$

Induced emf,
\[e=Bvl=4\times {{10}^{-5}}\times \frac{25}{3} \times 3=1\times {{10}^{-3}}V\]
That is, induced emf in vertical position is $1\times {{10}^{-3}}V$

Now considering the second case, that is when an athlete run with a rod in horizontal direction. Then the angle between the length of the rod and the magnetic field is zero.
That is $\sin\theta=\sin 0=0$
Therefore, the induced emf, $e=Blv\sin\theta= 0$

Since no magnetic field is cut when an athlete runs towards east with a rod in horizontal direction. According to Faraday’s law, induced emf is generated only when there is variation in magnetic or electric field with respect to time.

Therefore, there is an induced emf produced when an athlete runs towards east with rod in hold in vertical position and there is no induced emf produced when an athlete runs towards east with rod holding in horizontal position.

Therefore, the answer is option (B)

Note: While solving this question we need to take care of the direction of the rod. And also, we should know what is the direction of earth’s magnetic field. Don’t forget to convert velocity to m/s.