Answer to Question #129333 in Physics for Rimsha Ahmed

Question

A horseshoe magnet rests on a top-pan balance with a wire situated between the poles of the magnet.

With no current in the wire, the reading on the balance is 142.0 g.

With a current of 2.0 A in the wire in the direction XY (The wire is situated parallel to the poles of the magnet, XY is one direction the current can flow), the reading on the balance chages to 144.6 g.

What is the reading on the balance, when there is a current of 3.0 A in the wire in the direction YX (reverse of the direction of XY)?

A. 138.1 g
B. 140.7 g
C. 145.9 g
D. 148.5 g

Kindly, tell me how this question will be solved, please.

Thank you.

Answer 1

As per the given question,

When no current was flowing in the wire, in that situation, the weighting machine was reading 142.0 g

Now, for the second case, when the value of the current = 2A, in that case the weight of the wire gets increase, it means the direction of the magnetic force is downwards, so

$\Rightarrow(144.6-142)g= i_1BL$

$\Rightarrow 2.6g= 2BL ... (i)$

Here L is the length of the wire and B is the magnetic field of horseshoe magnet.

so, $BL=\frac{2.6 g}{2}$

Now the current in the wire gets 3A, then,

$\Delta W=3 BL = 3\times \frac{2.6g}{2} =3.9 g$

Hence machine will read $=(142+3.9 )gm =145.9 gm$