A 200-turn rectangular coil hacking dimension of 3.0 cm by 6.0 cm is place in a

Question

A 200-turn rectangular coil hacking dimension of 3.0 cm by 6.0 cm is place in a uniform magnetic field of magnitude 0.87T.
A) find the current in the coil if the maximum torque exerted on it by the magnetic field is 0.18N.m
B) find the magnitude of the torque on the coil when the magnetic field make an angle of 25wuth the normal to the plane of coil . What is the magnetic moment A 200-turn rectangular coil hacking dimension of 3.0 cm by 6.0 cm is place in a uniform magnetic field of magnitude 0.87T.
A) find the current in the coil if the maximum torque exerted on it by the magnetic field is 0.18N.m
B) find the magnitude of the torque on the coil when the magnetic field make an angle of 25wuth the normal to the plane of coil . What is the magnetic moment A 200-turn rectangular coil hacking dimension of 3.0 cm by 6.0 cm is place in a uniform magnetic field of magnitude 0.87T.
A) find the current in the coil if the maximum torque exerted on it by the magnetic field is 0.18N.m
B) find the magnitude of the torque on the coil when the magnetic field make an angle of 25wuth the normal to the plane of coil . What is the magnetic moment

Explanation / Answer

given that

magnetic field B = 0.87 T.

maximum torque t = 0.18 N*m

dimensions of coil

a= 3 cm = 0.03 m

b = 6 cm = 0.06 m

part (a)

we know that

torque t = m*B*sin (theta)

where m is megnetic moment.

& theta is angle b/w magnetic field and the normal to the plane of coil

given that torque is maximum so theta must be 90 deg (because sin 90 deg = 1)

so t = m*B

m = t / B

m = 0.18 / 0.87 = 0.20 A*m^2

we know that

m = I *A

where I is current and A is area of rectangular coil

A = a*b = 0.03 * 0.06 = 0.0018 m^2

I = m / A

I= 0.20 / 0.0018 = 114.94 A

part(b)

t = m*B*sin (theta)

given that theta = 25 degree

t = 0.20 * 0.87 * sin (25)

t = 0.20 * 0.87 * 0.42

t = 0.073 N*m

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