Name: Conversion of a Galvanometer Into a Voltmeter
Uploaded: 2019-07-19
Duration: 7 min 18 s
Description: This video helps us understand the working of a galvanometer. It also teaches us how to convert it into a voltmeter of the desired range both theoretically a...

Question

A $100$ turn rectangular coil $ABCD$ (in $XY$ plane) is hung from one arm of a balance (Fig). A mass $500 g$ is added to the other arm to balance the weight of the coil. A current $4.9 A$ passes through the coil and a constant magnetic field of $0.2 T$ acting inward (in xz plane) is switched on such that only arm $CD$ of length $1 cm$ lies in the field. How much additional mass $‘ m ’$ must be added to regain the balance?

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Accepted Answer

The magnetic field is perpendicular to arms $B C$ and $A D$ , so torque will act on $C D$ and $A B$ arms due to which the coil will rotate.
When current of $4.9 A$ pass through the coil, the balance measures mass of coil $500 g$ .
On arm $AD$ and $BC$ of rectangular coil magnetic force will be equal and opposite so it will rotate the coil horizontally not vertically up or down. So does not affect the balance.
When current of $4.9 A$ passes through the coil, downward force acts on arm $CD$ due to magnetic field. Length of arm $CD$ is $1 cm$ $(10^{- 2} m)$ .

Force acting on arm $C D$ is $F_{m} = I (B \times L) = B I L \sin θ$

$F_{m} = (0.2 T) (4.9 A) (10^{- 2} m) \sin 90 °$

$F_{m} = 0.98 \times 10^{- 2} N$

Now let the weight $m g$ is added on other side of beam balance to balance the coil

$m g = F_{m}$

$m (9.8 m s^{- 2}) = (0.98 \times 10^{- 2} N)$

$m = \frac{0.98}{9.8} \times 10^{- 2} kg = 10^{- 3} kg = 1 g$

Hence, $1 g$ of mass must be added to regain the balance.

NCERT Solutions for Chapter: Moving Charges and Magnetism, Exercise 4: LA

NCERT Physics Solutions for Exercise - NCERT Solutions for Chapter: Moving Charges and Magnetism, Exercise 4: LA

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