A long magnet is placed vertically with its S -pole resting on the table. A neutral point is obtained 10 cm from the pole due geographic north of it. If H=3.2 × 10-5 T, then the pole strength of magnet is

Question

A long magnet is placed vertically with its S -pole resting on the table. A neutral point is obtained 10 cm from the pole due geographic north of it. If H=3.2 × 10-5 T, then the pole strength of magnet is (A) 8 ab - A-cm -1 (B) 16 ab - A- cm -1 (C) 32 ab - A - cm -1 (D) 64 ab - A - cm -1

Tardigrade Admin · Accepted Answer

As, the magnet is long, we assume that the upper north pole produces no effect. But due to south pole of the magnet is equal and opposite to horizontal component of earth's magnetic field, i.e.,

B = (\frac{μ _{0}}{4 π}) \frac{m}{r ^{2}} = H

In CGS system,

\frac{μ _{0}}{4 π} = 1

∴ 1 \times \frac{m}{1 0 ^{2}} = 3.2 \times 1 0^{- 5} \times 1 0^{4}

(gauss)

m = 32 ab - am p - c m^{- 1}

Q. A long magnet is placed vertically with its $S$ -pole resting on the table. A neutral point is obtained $10 c m$ from the pole due geographic north of it. If $H = 3.2 \times 1 0^{- 5} T$ , then the pole strength of magnet is

$8 ab - A - c m^{- 1}$

$16 ab - A - c m^{- 1}$

$32 ab - A - c m^{- 1}$

$64 ab - A - c m^{- 1}$

Q. A long magnet is placed vertically with its S -pole resting on the table. A neutral point is obtained 10cm from the pole due geographic north of it. If H=3.2×10−5T, then the pole strength of magnet is

8ab−A−cm−1

16ab−A−cm−1

32ab−A−cm−1

64ab−A−cm−1

Q. A long magnet is placed vertically with its $S$ -pole resting on the table. A neutral point is obtained $10 c m$ from the pole due geographic north of it. If $H = 3.2 \times 1 0^{- 5} T$ , then the pole strength of magnet is

$8 ab - A - c m^{- 1}$

$16 ab - A - c m^{- 1}$

$32 ab - A - c m^{- 1}$

$64 ab - A - c m^{- 1}$