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GarryVolchara [31]
3 years ago
10

A single circular loop of wire of radius 0.45 m carries a constant current of 2.4 A. The loop may be rotated about an axis that

passes through the center and lies in the plane of the loop. When the orientation of the normal to the loop with respect to the direction of the magnetic field is 36 degrees, the torque on the coil is 1.5 N.m. What is the magnitude of the uniform magnetic field exerting this torque on the loop
Physics
1 answer:
Snowcat [4.5K]3 years ago
8 0

Answer:

The magnitude of the uniform magnetic field exerting this torque on the loop is 1.67 T

Explanation:

Given;

radius of the wire, r = 0.45 m

current on the loop, I = 2.4 A

angle of inclination, θ = 36⁰

torque on the coil, τ = 1.5 N.m

The torque on the coil is given by;

τ = NIBAsinθ

where;

B is the magnetic field

Area of the loop is given by;

A = πr² = π(0.45)² = 0.636 m

τ = NIBAsinθ

1.5 = (1 x 2.4 x 0.636 x sin36)B

1.5 = 0.8972B

B = 1.5 / 0.8972

B = 1.67 T

Therefore, the magnitude of the uniform magnetic field exerting this torque on the loop is 1.67 T

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Answer

given,

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   work done by the gravity = ?          

   work done by the tension = ?            

Work done by the gravity = - m g Δh            

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v_c is the velocity of the capsule

Since the total momentum must be conserved, we have

p_i = p_f = 0

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m_a v_a + m_c v_c=0

Solving the equation for v_c, we find

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(b) 520.8 N

We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:

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K=\frac{1}{2}mv^2

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