At a place the value of horizontal component of the earth's magnetic field H is 3 × 10-5 Wb / m 2. A metallic rod AB of length 2 m placed in east-west direction, having the end A towards east, falls vertically downward with a constant velocity of 50 m / s. Which end of the rod becomes positively charged and what is the value of induced potential difference between the two ends?

Question

At a place the value of horizontal component of the earth's magnetic field H is 3 × 10-5 Wb / m 2. A metallic rod AB of length 2 m placed in east-west direction, having the end A towards east, falls vertically downward with a constant velocity of 50 m / s. Which end of the rod becomes positively charged and what is the value of induced potential difference between the two ends? (A) End A, 3 × 10-3 mV (B) End A, 3 mV (C) End B, 3 × 10-3 mV (D) End B, 3 mV

Tardigrade Admin · Accepted Answer

Induced potential difference in the rod.

e = B v l

or

e = H v l

Given:

H = 3 \times 1 0^{- 5} Wb / m^{2}

,

v = 50 m / s

,

l = 2 m

∴ e = 3 \times 1 0^{- 5} \times 50 \times 2

= 3 \times 1 0^{- 3} V = 3 mV

Now, by Right hand palm rule.
If

H

is towards

N

and rod is moving downward,
then the current will move form

B

to

A

.
Hence, end

B

will become positive.

End $A, 3 \times 1 0^{- 3} mV$

End $A, 3 mV$

End $B, 3 \times 1 0^{- 3} mV$

End $B, 3 mV$

End A,3×10−3mV

End A,3mV

End B,3×10−3mV

End B,3mV

Induced potential difference in the rod. e=Bvl or e=Hvl Given: H=3×10−5Wb/m2, v=50m/s, l=2m ∴e=3×10−5×50×2 =3×10−3V=3mV Now, by Right hand palm rule. If H is towards N and rod is moving downward, then the current will move form B to A. Hence, end B will become positive.

End $A, 3 \times 1 0^{- 3} mV$

End $A, 3 mV$

End $B, 3 \times 1 0^{- 3} mV$

End $B, 3 mV$