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 current carrying loop consists of

3

identical quarter circles of radius

R

, lying in the positive quadrants of the

x - y, y - z

and

z - x

planes with their centres at the origin, joined together. Find the direction and magnitude of

B

at the origin.

Accepted Answer

Suppose, the magnetic field at the centre of origin due to current carrying loop is $B = \frac{μ_{\circ}}{4 π} \frac{I θ}{R}$ .

So, for the quarter of the loop, the loop subtends at angle $θ = \frac{π}{2}$ .

Thus, magnetic field due to quarter of the loop at the centre will be $B = \frac{μ_{\circ}}{8} \frac{I}{R}$ and the direction will perpendicular to the plane of quarter of the loop.

So for the given orientation the magnitude of the magnetic field at the centre of origin due to the quarter of loop in yz, zx and xy plane will be same but they will bw mutually perpendicular to each other i.e.

$B_{y z} = \frac{μ_{\circ}}{8} \frac{I}{R} \hat{i}$

$B_{z x} = \frac{μ_{\circ}}{8} \frac{I}{R} \hat{j}$

$B_{x y} = \frac{μ_{\circ}}{8} \frac{I}{R} \hat{k}$

Thus, net magnetic field is $B_{n e t} = B_{y z} + B_{z x} + B_{x y} = \frac{μ_{\circ}}{8} \frac{I}{R} \hat{i} + \frac{μ_{\circ}}{8} \frac{I}{R} \hat{j} + \frac{μ_{\circ}}{8} \frac{I}{R} \hat{k} = \frac{μ_{\circ}}{8} \frac{I}{R} (\hat{i} + \hat{j} + \hat{k})$

NCERT Solutions for Chapter: Moving Charges and Magnetism, Exercise 3: SA

NCERT Physics Solutions for Exercise - NCERT Solutions for Chapter: Moving Charges and Magnetism, Exercise 3: SA

Questions from NCERT Solutions for Chapter: Moving Charges and Magnetism, Exercise 3: SA with Hints & Solutions

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