Question

A short magnet of moment 6.75 Am2 produces a
neutral point on its axis. If horizontal component
of earth's magnetic field is 5 x 10-5 Wb/m, then
the distance of the neutral point should be
(a) 10 cm
(b) 20 cm
(c) 30 cm
(d) 40 cm​

Answer 1

Answer:

c

Explanation:

the earth magnet field magnitude on neutral point on axis

Answer 2

(c)30 cm

Explanation:

Magnetic moment=$m=6.75Am^2$

$B_H=5\times 10^{-5}Wb/m^2$

We know that

$B_H=\frac{\mu_0}{4\pi}(\frac{2m}{r^3})$

Where $\frac{\mu_0}{4\pi}=10^{-7}H/m$

m=Magnetic moment

Substitute the values

$5\times 10^{-5}=\frac{10^{-7}\times 2\times 6.75}{r^3}$

$r^3=\frac{10^{-7}\times 2\times 6.75}{5\times 10^{-5}}$

$r=(\frac{10^{-7}\times 2\times 6.75}{5\times 10^{-5}})^{\frac{1}{3}$

$r=0.3 m$

1 m=100 cm

$r=0.3\times 100=30 cm$

Hence, option c is true.

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https://brainly.in/question/14612552:Answered by Abhishek