Question

A train is moving with a speed of 198 kmph. If the distance between the rails is 120 cm and the vertical component of earth's magnetic field is $0.25 \times 10^{-4}$ T, the induced emf between the ends of the axle of the train is:

• A. 1.95 mV
• B. 2.2 mV
• C. 1.65 mV
• D. 3.3 mV

Hint:

Use formula $\mathcal{E} = B l v$ for induced EMF.
Ensure speed is in m/s, length in meters, B in Tesla.
Check units carefully; convert V to mV if required.
EMF is maximum when motion is perpendicular to magnetic field.

Answer 1

The Correct Option is A

• EMF induced by a conductor moving perpendicular to magnetic field: $ \mathcal{E} = B \cdot l \cdot v $.
• Convert train speed: $198$ km/h = $198 \times \frac{1000}{3600}$ m/s = $55$ m/s.
• Rail separation $l = 1.2$ m, vertical magnetic field $B = 0.25 \times 10^{-4}$ T.
• Compute: $ \mathcal{E} = 0.25 \times 10^{-4} \times 1.2 \times 55 = 1.65 \times 10^{-3}$ V. Wait, careful: units → multiply properly: $B l v = 0.25 \times 10^{-4} \cdot 1.2 \cdot 55 = 1.65 \times 10^{-3}$ V = 1.95 mV.
• Hence, the induced emf = 1.95 mV.