Calculate the induced emf between the ends of an axle of a railway carriage 1.75 m long traveling on level ground with a uniform velocity of 50 kmph. The vertical component of Earth's magnetic field (B_{v}) is 5 × 10^{-5} T.

#### Solution 1

**Data:** l = 1.75 m, v = 50 km/h = `50 xx 5/18` m/s,

B_{v} = 5 × 10^{-5} T

The area swept out by the wing per unit time= lv.

∴ The magnetic flux cut by the wing per unit time

`= ("d"phi_"m")/"dt" = "B"_"v" ("lv")`

`= (5 xx 10^-5)(1.75)(50 xx 5/18) = 121.5 xx 10^-5`

= 1.215 m Wb/s

Therefore, the magnitude of the induced emf,

|e| = 1.215 mV

#### Solution 2

**Given: **

l = 1.75 m, B_{V} = 5 × 10^{-5} T,

v = 50 km/h = 50 × `5/18` m/s

**To find: **Induced emf

**Formula: **e = Blv

**Calculation: **

From formula

e = `5 xx 10^-5 xx 1.75 xx 50 xx 5/18`

= `121.5 xx 10^-5` V

**= 1.215 mV**

**The induced emf is 1.215 mV. **