Figure shows a metallic wire of resistance 0.20 Ω sliding on a horizontal, U-shaped metallic rail. The separation between the parallel arms is 20 cm. An electric current of 2.0 μA passes through the wire when it is slid at a rate of 20 cm s–1. If the horizontal component of the earth’s magnetic field is 3.0 × 10–5 T, calculate the dip at the place.


Given:


Resistance of wire(R) = 0.2 Ω


Length of wire(l) = 20 cm = 0.2 m


Current(I) = 2 μA = 2 x 10-6 A


Velocity with which the wire moves(v) = 20 cms-1 = 0.2 ms-1


Horizontal component of earth’s magnetic field(BH) = 3.0 × 10–5 T


Formula used:


Angle of dip ) … (i), where BV = vertical component of earth’s magnetic field, BH = horizontal component of earth’s magnetic field.


Now, emf induced in the wire … (ii), where BV = vertical component of earth’s magnetic field, l = length of wire, v = velocity with which the wire moves


Also, by Ohm’s law … (iii), where E = emf, I = current, v = velocity.


Equating (ii) and (iii) we get


= IR ⇒


Substituting the given values, we get


= 10-5 T


Hence, angle of dip ⇒ (Ans)


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