Question

Two identical short bar magnets each of magnetic moment $10A{m}^2$ are placed at a separation of $10cm$ between their centres such that their axes are perpendicular to each other. The magnetic field at a point midway between the two magnets is

• A. $\sqrt{5}\times {10}^{-6}T$
• B. $1.8\times {10}^{-2}T$
• C. $4\times {10}^{4}T$
• D. $12\times {10}^{-4}T$

Answer 1

Answer: (B)

$1.8\times {10}^{-2}T$

Given: M=10A$m^2$

Solution: When two identical short bar magnets are placed such that their axes are perpendicular to each other then that point will be axial position for first bar magnet and equatorial for another.

Formula used for magnetic field at axial position will be:

$B_1=\frac{{\mu }_0}{4\pi }\times \frac{2M}{r^3}$

Now, put values in the formula

$B_1={10}^{-7}\times \frac{2\times 10}{{\left(0.05\right)}^3}$=0.016

Formula used for magnetic field at equatorial position will be:

$B_2=\frac{{\mu }_0}{4\pi }\times \frac{M}{r^3}$

Now, put values in the formula

$B_2={10}^{-7}\times \frac{10}{{\left(0.05\right)}^3}$=0.008

Net magnetic field=$\sqrt{B_1^2+B_2^2}$

B=$1.8\times {10}^{-2}$