# 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 componen - Physics

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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 (Bv) is 5 × 10-5 T.

#### Solution 1

Data: l = 1.75 m, v = 50 km/h = 50 xx 5/18 m/s,
Bv = 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, BV = 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.

Concept: Induced Emf in a Stationary Coil in a Changing Magnetic Field
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#### APPEARS IN

Balbharati Physics 12th Standard HSC Maharashtra State Board
Chapter 12 Electromagnetic induction
Exercises | Q 11 | Page 287