Question

A coil of 100 tums is wound uniformly over a iron ring of 0.5 meter length and carry a current of 1 Amp, what will be the magnetic
the magnet
Options
O 20 AT/m
O 50 AT/m
O 100 AT/M
0 200 AT/M​

Answer 1

Answer:

N=100turns

d=0.2m

R=

2

d

=

2

0.2

=0.1m

I=1A

B=(

2R

μ

o

I

)N

=

2×0.1

4π×10

−7

×1

×1000

=2π×10

−4

Tesla

Answer 2

Answer:

Option d will be correct answer as $H = 200 A/m$

Explanation:

we know that the Magnetic field strength, H

H =$\frac{NXI}{L}$

where N =  THe Numbers of turn Of coil

Given that N = 100

I = The current carried by the iron ring=

Given that I= 1 Amp

L=The length of the iron ring

That  is given = .5 Meter

Now we have to find out the Magnetic field strength, H

by using formula

$H=\frac{NXI}{L}$

Putting the value of N , I, L from the above

$H=\frac{100X1}{.5}$

$H = 200 A/m$

Hence, the magnetic field strength $H = 200 A/m$

Option d will be correct answer as $H = 200 A/m$

#SPJ3