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tensa zangetsu [6.8K]

3 years ago

14

A 0.12-kg metal rod carrying a current of current 4.1 A glides on two horizontal rails separation 6.3 m apart. If the coefficien

t of kinetic friction between the rod and rails is 0.18 and the kinetic friction force is 0.212 N , what vertical magnetic field is required to keep the rod moving at a constant speed of 5.1 m/s

1 answer:

Neporo4naja [7]3 years ago

7 0

Answer:

The magnetic field is $B = 8.20 *10^{-3} \ T$

Explanation:

From the question we are told that

The mass of the metal rod is $m = 0.12 \ kg$

The current on the rod is $I = 4.1 \ A$

The distance of separation(equivalent to length of the rod ) is $L = 6.3 \ m$

The coefficient of kinetic friction is $\mu_k = 0.18$

The kinetic frictional force is $F_k = 0.212 \ N$

The constant speed is $v = 5.1 \ m/s$

Generally the magnetic force on the rod is mathematically represented as

$F = B * I * L$

For the rod to move with a constant velocity the magnetic force must be equal to the kinetic frictional force so

$F_ k = B* I * L$

=> $B = \frac{F_k}{L * I }$

=> $B = \frac{0.212}{ 6.3 * 4.1 }$

=> $B = 8.20 *10^{-3} \ T$

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