Question

The magnetic moment of a magnet is 0.1 amp × m². It is suspended in a magnetic field of intensity 3 × 10⁻⁴ weber/m². The couple acting upon it when deflected by 30° from the magnetic field is(a) 1 × 10⁻⁵ N m(b) 1.5 × 10⁻⁵ N m(c) 2 × 10⁻⁵ N m(d) 2.5 × 10⁻⁵ N m

Answer 1

Answer:

option a

Explanation:

GIVEN

M=0.1 amp×m²

B=3×10^-4

TO FIND

TORQUE

SOLUTION

$torque = mb \sin( theta)$

Answer 2

Answer:

option b).1.5 × 10⁻⁵ N m

Explanation:

we know that

M=0.1

B=3×10⁻⁴

then

T=M×B×sin (theta)

T=0.1×3×10⁻⁴×sin 30

=0.1×3×10⁻⁴×0.5

=0.1×0.5× 10⁻⁴

=1.5×10⁻⁵

therefore ,couple aganist magnetic field isT=1.5 × 10⁻⁵

T=1.5 × 10⁻⁵ that is option b).1.5 × 10⁻⁵ N m

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