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Nezavi [6.7K]
3 years ago
13

A wire is formed into a circle having a diameter of 10.3 cm and is placed in a uniform magnetic field of 2.98 mT. The wire carri

es a current of 5.00 A. Find the maximum torque on the wire.
Physics
1 answer:
Ksju [112]3 years ago
8 0

Answer:

T(max) = 1.17 × 10⁻⁴Nm

= 117μNm

Explanation:

T = BIA sinθ

A = area enclosed

θ = angle between normal plane

for max. torque θ = 90, (sin90° =1)

T = BIA sin90°

T= BI (πd/4)

T = T_m_a_x = \frac{1}{4} (2.98 * 10^-^3)(5)(\pi )\\T_m_a_x = 1.17 * 10^-^4Nm\\T_m_a_x = 117UNm

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A circular loop of flexible iron wire has an initial circumference of 165cm , but its circumference is decreasing at a constant
ArbitrLikvidat [17]

Answer:

emf induced is 0.005445 V and direction is clockwise because we can see area is decrease and so that flux also decrease so using right hand rule direction of current here clockwise

Explanation:

Given data

initial circumference = 165 cm

rate = 12.0 cm/s

magnitude = 0.500 T

tome = 9 sec

to find out

emf induced and direction

solution

we know emf in loop is - d∅/dt    ........1

here ∅ = ( BAcosθ)

so we say angle is zero degree and magnetic filed is uniform here so that

emf = - d ( BAcos0) /dt

emf = - B dA /dt     ..............2

so  area will be

dA/dt = d(πr²) / dt

dA/dt = 2πr dr/dt

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emf = - B 2πr dr/dt    ................3

and

here we find rate of change of radius that is

dr/dt = 12/2π = 1.91  10^{-2}cm/s

so when 9.0s have passed that radius of coil = 26.27 - 191 (9)

radius = 9.08 10^{-2} cm

so now from equation 3 we find emf

emf = - (0.500 )  2π(9.08 10^{-2} )   1.91  10^{-2}

emf = - 0.005445

and magnitude of emf = 0.005445 V

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